Trigonometry

1. Trigonometry  Trigonometry is derived from Greek words trigonon (three angles) and metron ( 
measure).  Trigonometry is the branch of mathematics which deals with triangles, particularly 
triangles in a plane where one angle of the triangle is 90 degrees  Triangles on a sphere are also 
studied, in spherical trigonometry.  Trigonometry specifically deals with the relationships between 
the sides and the angles of triangles, that is, on the trigonometric functions, and with calculations 
based on these functions. 
2. HISTORY  The origins of trigonometry can be traced to the civilizations of ancient Egypt, 
Mesopotamia and the Indus Valley, more than 4000 years ago.  Some experts believe that 
trigonometry was originally invented to calculate sundials, a traditional exercise in the oldest books. 
 The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus circa 
150 BC, who compiled a trigonometric table using the sine for solving triangles.  The Sulba Sutras 
written in India, between 800 BC and 500 BC, correctly compute the sine of π/4 (45°) as 1/√2 in a 
procedure for circling the square (the opposite of squaring the circle).  Many ancient 
mathematicians like Aryabhata, Brahmagupta,Ibn Yunus and Al-Kashi made significant contributions 
in this field(trigonometry). 
3. RIGHT TRIANGLE  A triangle in which one angle is equal to 90 is called right triangle.  The side 
opposite to the right angle is known as hypotenuse.  AB is the hypotenuse  The other two sides 
are known as legs.  AC and BC are the legs Trigonometry deals with Right Triangles Only 
4. PYTHAGORAS THEOREM In any right triangle, the area of the square whose side is the 
hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. In the figure 
AB2 = BC2 + AC2 5 
5. Sine(sin) TRIGONOMETRIC RATIOS opposite side/hypotenuse Cosine(cos) adjacent 
side/hypotenuse Tangent(tan) opposite side/adjacent side Cosecant(cosec) hypotenuse/opposite 
side Secant(sec) hypotenuse/adjacent side Cotangent(cot) adjacent side/opposite side 
6. Simple Way of Remembering Trigonometric Ratios Sine(sin) Cosine(cos) Tangent(tan) Opposite (O) 
Adjacent (A) Opposite (O) Hypotenuse (H) Hypotenuse (H) Adjacent (A) Cosecant(cosec) 
Secant(sec) Cotangent(cot) 
7. VALUES OF TRIGONOMETRIC FUNCTION OF ANGLE A sin = a/c cos = b/c tan = a/b cosec 
= c/a sec = c/b cot = b/a 
8. VALUES OF TRIGONOMETRIC FUNCTION 0 30 45 Sine 0 1/2 1/ 2 Cosine 1 3/2 Tangent 0 1/ 3 
Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 1/ 3 0 3 1/ 2 60 
3/2 1/2 1 1 3 90 1 0 Not defined 
9. TRIGONOMETRIC IDENTITIES 2 • sin2A + cos A = 1 • 1 + tan2A = sec2A • 1 + cot2A = cosec2A • 
sin(A/2) = {(1-cosA)/2} • Cos(A/2)= {(1+cosA)/2} • Tan(A/2)= {(1-cosA)/(1+cosA)} 
10. ANGLE IN STANDARD POSITION 
11. FORMULAS • sin (A+B) = sin A cos B + cos A sin B • sin (A-B) = sin A cos B - cos A sin B • cos 
(A+B) = cos A cos B - sin A sin B • cos(A-B) = cos A cos B + sin A sin B • tan (A+B) = [tan A + tan B] 
/ [1 - tan A tan B] • tan (A-B) = [tan A - tan B] / [1 + tan A tan B] 
12. • APPLICATIONS OF TRIGONOMETRY This field of mathematics can be applied in 
astronomy,navigation, music theory, acoustics, optics, analysis of financial markets, electronics, 
probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy, 
chemistry, number theory (and hence cryptology), seismology, meteorology, oceanography, many
physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical 
engineering, mechanical engineering, civil engineering, computer graphics, cartography, 
crystallography and game development. 
13. APPLICATIONS OF TRIGONOMETRY IN ASTRONOMY • Since ancient times trigonometry was 
used in astronomy. • The technique of triangulation is used to measure the distance to nearby stars. 
• In 240 B.C., a mathematician named Eratosthenes discovered the radius of the Earth using 
trigonometry and geometry. • In 2001, a group of European astronomers did an experiment that 
started in 1997 about the distance of Venus from the Sun. Venus was about 105,000,000 kilometers 
away from the Sun . 
14. APPLICATION OF TRIGONOMETRY IN ARCHITECTURE • Many modern buildings have 
beautifully curved surfaces. • Making these curves out of steel, stone, concrete or glass is extremely 
difficult, if not impossible. • One way around to address this problem is to piece the surface together 
out of many flat panels, each sitting at an angle to the one next to it, so that all together they create 
what looks like a curved surface. • • The more regular these shapes, the easier the building process. 
Regular flat shapes like squares, pentagons and hexagons, can be made out of triangles, and so 
trigonometry plays an important role in architecture. 
15. WAVES • The graphs of the functions sin(x) and cos(x) look like waves. Sound travels in waves, 
although these are not necessarily as regular as those of the sine and cosine functions. • However, a 
few hundred years ago, mathematicians realized that any wave at all is made up of sine and cosine 
waves. This fact lies at the heart of computer music. • Since a computer cannot listen to music as we 
do, the only way to get music into a computer is to represent it mathematically by its constituent 
sound waves. • This is why sound engineers, those who research and develop the newest advances 
in computer music technology, and sometimes even composers have to understand the basic laws 
of trigonometry. • Waves move across the oceans, earthquakes produce shock waves and light can 
be thought of as traveling in waves. This is why trigonometry is also used in oceanography, 
seismology, optics and many other fields like meteorology and the physical sciences. 
16. DIGITAL IMAGING • In theory, the computer needs an infinite amount of information to do this: it 
needs to know the precise location and colour of each of the infinitely many points on the image to 
be produced. In practice, this is of course impossible, a computer can only store a finite amount of 
information. • To make the image as detailed and accurate as possible, computer graphic designers 
resort to a technique called triangulation. • As in the architecture example given, they approximate 
the image by a large number of triangles, so the computer only needs to store a finite amount of 
data. • The edges of these triangles form what looks like a wire frame of the object in the image. 
Using this wire frame, it is also possible to make the object move realistically. • Digital imaging is 
also used extensively in medicine, for example in CAT and MRI scans. Again, triangulation is used 
to build accurate images from a finite amount of information. • It is also used to build "maps" of 
things like tumors, which help decide how x-rays should be fired at it in order to destroy it. 
17. Finally, Trigonometry is a branch of Mathematics with several important and useful applications. 
Hence it attracts more and more research with several theories published year after year 
USE OF TRIGONOMETRY IN REAL LIFE SUPREIYA CLASS : X - A 
he word “Trigonometry” comes from the Greek words: ‘Trigonon’ meaning 
mathematics which deals with the measurement of the sides and the angles of a triangle and the 
problems allied with angles.
ORIGIN OF ‘SINE’ “Trigonometry is not the work of any one person or nation. Its history spans thousands 
it toda -jya ’ 
mund Gunter (1581-1626) first used the abbreviated 
notation ‘sin ’ . Aryabhata A.D. 476-550 
function arose from the need to compute the sine of the complementar 
Jonas Moore first used the abbreviated notation ‘cos’ Edmund Gunter (1581 –1626) 
THE TRIGONOMETRIC RATIOS Abbr. Descriptio n Sine sin Opposite Cosine cos Tangent tan Cotangent 
The Cosecant, Secant, and Cotangent are the Reciprocals of the Sine, Cosine,and Tangent respectively . 
Function cot Secant Note: The formulas provided are in respect to the picture. sec Cosecan cosec t 
Hypotenus e Adjacent Hypotenus e Opposite Adjacent Adjacent Opposite Hypotenus e Adjacent 
Hypotenus e 
THE TRIGONOMETRIC VALUES Angle A 0o 30o 45o 60o 90o sin A 0 1 1 √3 1 1 2 √3 √2 1 2 1 0 0 2 1 √2 1 2 
√3 √3 2 √2 2 2 √2 √3 2 √3 √3 1 1 cos A tan A cosec A sec A cot A Not Defined 1 Not Defined √3 Not 
Defined 1 Not Defined 0 
nometric 
your room !! 
Perpendicular = 6 mts. Hypotenuse = 12mts. 
= Sin 30 o Angle of Elevation = 30o 
1. THANK YTrigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 
2.  The word ‘trigonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ (meaning sides) 
and ‘metron’ (meaning measure).  Trigonometry is the study of relationships between the sides and 
angles of a triangle.  Early astronomers used it to find out the distances of the stars and planets 
from the Earth.  Even today, most of the technologically advanced methods used in Engineering
and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights 
Reserved 
3.  A triangle in which one angle is equal to 90 is called right triangle.  The side opposite to the right 
angle is known as hypotenuse. AC is the hypotenuse  The other two sides are known as legs. AB 
and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights 
Reserved 
4.  In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of 
areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = (Perpendicular)2 + 
(Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 
5. Pythagoras Theorem Proof:  Given: Δ ABC is a right angled triangle where B = 900 And AB = P, 
BC= b and AC = h.  To Prove: h2 = p2 + b2  Construction : Draw a BD from B to AC , where AD = 
x and CB = h-x ,  Proof : In Δ ABC and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ ABC and Δ BDC both 
are similar So by these similarity, p b h A B C 
6. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x) 
Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A 
B C 
7.  Let us take a right triangle ABC  Here, ∠ ACB ( ) is an acute angle.  The position of the side AB 
with respect to angle . We call it the side opposite to angle .  AC is the hypotenuse of the right 
triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB 
Sideoppositetoangle Side adjacent to angle ‘ ’ © iTuto r. 2000-2013. All Rights Reserved 
8.  The trigonometric ratios of the angle C in right ABC as follows :  Sine of C = =  Cosine of C= = A 
CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side 
adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 
9.  Tangent of C = =  Cosecant of C= =  Secant of C = A CB Sideoppositetoangle Side adjacent to 
angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite 
to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved 
10.  Cotangent of C  Above Trigonometric Ratio arbitrates as sin C, cos C, tan C , cosec C , sec C, 
Cot C .  If the measure of angle C is ‘ ’ then the ratios are : sin , cos , tan , cosec , sec and cot A CB 
Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = © 
iTutor. 2000-2013. All Rights Reserved 
11.  Tan =  Cosec = 1 / Sin  Sec = 1 / Cos  Cot = Cos / Sin = 1 / Tan A CB p b h © iTutor. 2000- 
2013. All Rights Reserved cos sin 
12. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h © 
iTutor. 2000-2013. All Rights Reserved 
13. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is 
also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras 
Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000 - 
2013. All Rights Reserved 
14.  Sin 450 = = = = 1/ 2  Cos 450 = = = = 1/ 2  Tan 450 = = = = 1  Cosec 450 = 1 / sin 450 = 1 / 1/ 
2 = 2  Sec 450 = 1 / cos 450 = 1 / 1/ 2 = 2  Cot 450 = 1 / tan 450 = 1 / 1 = 1 Side opposite to 450
Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side 
adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved 
15.  Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore, 
∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD ----- 
---- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) Now observe that: Δ ABD is 
a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor. 
2000-2013. All Rights Reserved 
16.  As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the 
triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD 
= a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to 
300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 
17. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 = 
1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric rat ios of 600 AD AB a 3 2a BD 
AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 
18. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec 
600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a 
600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 
19. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined 
Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 © 
iTutor. 2000-2013. All Rights Reserved 
20.  Relation of with Sin when 00 900 The greater the value of ‘ ’, the greater is the value of Sin . 
Smallest value of Sin = 0 Greatest value of Sin = 1  Relation of with Cos when 00 900 The 
greater the value of ‘ ’, the smaller is the value of Cos . Smallest value of Cos = 0 Greatest value 
of Cos = 1 © iTutor. 2000-2013. All Rights Reserved 
21.  Relation of with tan when 00 900 Tan increases as ‘ ’ increases But ,tan is not defined at ‘ ’ = 
900 Smallest value of tan = 0 © iTutor. 2000-2013. All Rights Reserved 
22.  If 00 900 1. Sin(900- ) = Cos 2. Cos(900- ) = Sin  If 00< 900 1. Tan(900- ) = Cot 2. Sec(900- ) = 
Cosec  If 00 < 900 1. Cot(900- )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All 
Rights Reserved 
23. Sin2 +Cos2 = 1 Sec2 -Tan2 = 1 Cosec2 - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved 
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25. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 
26. 
relationships between the sides and ang 
technologically advanced methods used in Engineering and Physical Sciences are based on 
trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 
27.
known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © 
iTutor. 2000-2013. All Rights Reserved 
28. 
sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = 
(Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 
29. 
rom B to AC 
, where AD = x and CB = h- --------(AA) In Δ 
ABC and Δ BDC both are similar So by these similarity, p b h A B C 
30. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h 
(h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h 
p And p b h A B C 
31. ∠ 
side AB with respect to angle . We call it the si 
of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A 
CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved 
32. 
of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse 
AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 
33. 
adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C 
Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights 
Reserved 
34. 
sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side 
adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 
35. 
2000-2013. All Rights Reserved cos sin 
36. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p 
b h © iTutor. 2000-2013. All Rights Reserved 
37. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other 
angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by 
Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 
450 © iTutor. 2000-2013. All Rights Reserved 
38. 
450 = 1 / 1/ 2 
Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a 
Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights 
Reserved
39. 
therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ 
ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) 
Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 
60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved 
40. he sides 
of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - 
(a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a 
a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 
41. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / 
cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 
AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights 
Reserved 
42. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 
3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a 
AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 
43. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not 
defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 
3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 
44. 
-2013. All Rights Reserved 
45. 
2000-2013. All Rights Reserved 
46. - ) = Cos 2. Cos(900- - ) = Cot 2. 
Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h © 
iTutor. 2000-2013. All Rights Reserved 
47. - - Cot2 = 1 © iTutor. 2000-2013. All Rights 
Reserved 
48. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 
49. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 
50. gonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ 
out the dist 
technologically advanced methods used in Engineering and Physical Sciences are based on 
trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 
51. 
known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © 
iTutor. 2000-2013. All Rights Reserved
52. 
sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = 
(Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 
53. 
, where AD = x and CB = h- C and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ 
ABC and Δ BDC both are similar So by these similarity, p b h A B C 
54. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h 
(h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h 
p And p b h A B C 
55. ∠ 
of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A 
CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved 
56. 
of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse 
AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 
57. tetoangle Side 
adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C 
Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights 
Reserved 
58. in C, cos C, tan C , cosec C , 
sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side 
adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 
59. 
2000-2013. All Rights Reserved cos sin 
60. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p 
b h © iTutor. 2000-2013. All Rights Reserved 
61. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other 
angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by 
Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 
450 © iTutor. 2000-2013. All Rights Reserved 
62. 
tan 450 = 1 / 1 = 1 
Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a 
Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights 
Reserved 
63. le in an equilateral triangle is 60°, 
therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ 
ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT)
Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 
60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved 
64. 
of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - 
(a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a 
a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 
65. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / 
cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 
AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights 
Reserved 
66. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 
3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a 
AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 
67. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not 
defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 
3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 
68. greater the value of ‘ ’, the greater is the value of 
e of Cos = 1 © iTutor. 2000-2013. All Rights Reserved 
69. 
-2013. All Rights Reserved 
70. - ) = Cos 2. Cos(900- - ) = Cot 2. 
Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h © 
iTutor. 2000-2013. All Rights Reserved 
71. - - Cot2 = 1 © iTutor. 2000-2013. All Rights 
Reserved 
72. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 
73. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 
74. ee), ‘gon’ 
most of the 
technologically advanced methods used in Engineering and Physical Sciences are based on 
trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 
75. site to 
known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © 
iTutor. 2000-2013. All Rights Reserved 
76. any right triangle, the area of the square whose side is the hypotenuse is equal to the 
sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 =
(Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 
77. Pyt 
, where AD = x and CB = h- --------(AA) In Δ 
ABC and Δ BDC both are similar So by these similarity, p b h A B C 
78. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h 
(h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h 
p And p b h A B C 
79. ∠ 
of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A 
CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved 
80. 
of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse 
AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 
81. 
adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C 
Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights 
Reserved 
82. 
sec 
sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side 
adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 
83. 
2000-2013. All Rights Reserved cos sin 
84. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p 
b h © iTutor. 2000-2013. All Rights Reserved 
85. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other 
angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by 
Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 
450 © iTutor. 2000-2013. All Rights Reserved 
86. 
Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a 
Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights 
Reserved 
87. °, 
therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ 
ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) 
Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 
60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved
88. 
of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - 
(a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a 
a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 
89. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / 
cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 
AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights 
Reserved 
90. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 
3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a 
AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 
91. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not 
defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 
3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 
92. ter is the value of 
-2013. All Rights Reserved 
93. 
-2013. All Rights Reserved 
94. - ) = Cos 2. Cos(900- 1. Tan(900- ) = Cot 2. 
Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h © 
iTutor. 2000-2013. All Rights Reserved 
95. - - Cot2 = 1 © iTutor. 2000-2013. All Rights 
Reserved 
96. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 
97.

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Trigonometry

  • 1. 1. Trigonometry  Trigonometry is derived from Greek words trigonon (three angles) and metron ( measure).  Trigonometry is the branch of mathematics which deals with triangles, particularly triangles in a plane where one angle of the triangle is 90 degrees  Triangles on a sphere are also studied, in spherical trigonometry.  Trigonometry specifically deals with the relationships between the sides and the angles of triangles, that is, on the trigonometric functions, and with calculations based on these functions. 2. HISTORY  The origins of trigonometry can be traced to the civilizations of ancient Egypt, Mesopotamia and the Indus Valley, more than 4000 years ago.  Some experts believe that trigonometry was originally invented to calculate sundials, a traditional exercise in the oldest books.  The first recorded use of trigonometry came from the Hellenistic mathematician Hipparchus circa 150 BC, who compiled a trigonometric table using the sine for solving triangles.  The Sulba Sutras written in India, between 800 BC and 500 BC, correctly compute the sine of π/4 (45°) as 1/√2 in a procedure for circling the square (the opposite of squaring the circle).  Many ancient mathematicians like Aryabhata, Brahmagupta,Ibn Yunus and Al-Kashi made significant contributions in this field(trigonometry). 3. RIGHT TRIANGLE  A triangle in which one angle is equal to 90 is called right triangle.  The side opposite to the right angle is known as hypotenuse.  AB is the hypotenuse  The other two sides are known as legs.  AC and BC are the legs Trigonometry deals with Right Triangles Only 4. PYTHAGORAS THEOREM In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. In the figure AB2 = BC2 + AC2 5 5. Sine(sin) TRIGONOMETRIC RATIOS opposite side/hypotenuse Cosine(cos) adjacent side/hypotenuse Tangent(tan) opposite side/adjacent side Cosecant(cosec) hypotenuse/opposite side Secant(sec) hypotenuse/adjacent side Cotangent(cot) adjacent side/opposite side 6. Simple Way of Remembering Trigonometric Ratios Sine(sin) Cosine(cos) Tangent(tan) Opposite (O) Adjacent (A) Opposite (O) Hypotenuse (H) Hypotenuse (H) Adjacent (A) Cosecant(cosec) Secant(sec) Cotangent(cot) 7. VALUES OF TRIGONOMETRIC FUNCTION OF ANGLE A sin = a/c cos = b/c tan = a/b cosec = c/a sec = c/b cot = b/a 8. VALUES OF TRIGONOMETRIC FUNCTION 0 30 45 Sine 0 1/2 1/ 2 Cosine 1 3/2 Tangent 0 1/ 3 Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 1/ 3 0 3 1/ 2 60 3/2 1/2 1 1 3 90 1 0 Not defined 9. TRIGONOMETRIC IDENTITIES 2 • sin2A + cos A = 1 • 1 + tan2A = sec2A • 1 + cot2A = cosec2A • sin(A/2) = {(1-cosA)/2} • Cos(A/2)= {(1+cosA)/2} • Tan(A/2)= {(1-cosA)/(1+cosA)} 10. ANGLE IN STANDARD POSITION 11. FORMULAS • sin (A+B) = sin A cos B + cos A sin B • sin (A-B) = sin A cos B - cos A sin B • cos (A+B) = cos A cos B - sin A sin B • cos(A-B) = cos A cos B + sin A sin B • tan (A+B) = [tan A + tan B] / [1 - tan A tan B] • tan (A-B) = [tan A - tan B] / [1 + tan A tan B] 12. • APPLICATIONS OF TRIGONOMETRY This field of mathematics can be applied in astronomy,navigation, music theory, acoustics, optics, analysis of financial markets, electronics, probability theory, statistics, biology, medical imaging (CAT scans and ultrasound), pharmacy, chemistry, number theory (and hence cryptology), seismology, meteorology, oceanography, many
  • 2. physical sciences, land surveying and geodesy, architecture, phonetics, economics, electrical engineering, mechanical engineering, civil engineering, computer graphics, cartography, crystallography and game development. 13. APPLICATIONS OF TRIGONOMETRY IN ASTRONOMY • Since ancient times trigonometry was used in astronomy. • The technique of triangulation is used to measure the distance to nearby stars. • In 240 B.C., a mathematician named Eratosthenes discovered the radius of the Earth using trigonometry and geometry. • In 2001, a group of European astronomers did an experiment that started in 1997 about the distance of Venus from the Sun. Venus was about 105,000,000 kilometers away from the Sun . 14. APPLICATION OF TRIGONOMETRY IN ARCHITECTURE • Many modern buildings have beautifully curved surfaces. • Making these curves out of steel, stone, concrete or glass is extremely difficult, if not impossible. • One way around to address this problem is to piece the surface together out of many flat panels, each sitting at an angle to the one next to it, so that all together they create what looks like a curved surface. • • The more regular these shapes, the easier the building process. Regular flat shapes like squares, pentagons and hexagons, can be made out of triangles, and so trigonometry plays an important role in architecture. 15. WAVES • The graphs of the functions sin(x) and cos(x) look like waves. Sound travels in waves, although these are not necessarily as regular as those of the sine and cosine functions. • However, a few hundred years ago, mathematicians realized that any wave at all is made up of sine and cosine waves. This fact lies at the heart of computer music. • Since a computer cannot listen to music as we do, the only way to get music into a computer is to represent it mathematically by its constituent sound waves. • This is why sound engineers, those who research and develop the newest advances in computer music technology, and sometimes even composers have to understand the basic laws of trigonometry. • Waves move across the oceans, earthquakes produce shock waves and light can be thought of as traveling in waves. This is why trigonometry is also used in oceanography, seismology, optics and many other fields like meteorology and the physical sciences. 16. DIGITAL IMAGING • In theory, the computer needs an infinite amount of information to do this: it needs to know the precise location and colour of each of the infinitely many points on the image to be produced. In practice, this is of course impossible, a computer can only store a finite amount of information. • To make the image as detailed and accurate as possible, computer graphic designers resort to a technique called triangulation. • As in the architecture example given, they approximate the image by a large number of triangles, so the computer only needs to store a finite amount of data. • The edges of these triangles form what looks like a wire frame of the object in the image. Using this wire frame, it is also possible to make the object move realistically. • Digital imaging is also used extensively in medicine, for example in CAT and MRI scans. Again, triangulation is used to build accurate images from a finite amount of information. • It is also used to build "maps" of things like tumors, which help decide how x-rays should be fired at it in order to destroy it. 17. Finally, Trigonometry is a branch of Mathematics with several important and useful applications. Hence it attracts more and more research with several theories published year after year USE OF TRIGONOMETRY IN REAL LIFE SUPREIYA CLASS : X - A he word “Trigonometry” comes from the Greek words: ‘Trigonon’ meaning mathematics which deals with the measurement of the sides and the angles of a triangle and the problems allied with angles.
  • 3. ORIGIN OF ‘SINE’ “Trigonometry is not the work of any one person or nation. Its history spans thousands it toda -jya ’ mund Gunter (1581-1626) first used the abbreviated notation ‘sin ’ . Aryabhata A.D. 476-550 function arose from the need to compute the sine of the complementar Jonas Moore first used the abbreviated notation ‘cos’ Edmund Gunter (1581 –1626) THE TRIGONOMETRIC RATIOS Abbr. Descriptio n Sine sin Opposite Cosine cos Tangent tan Cotangent The Cosecant, Secant, and Cotangent are the Reciprocals of the Sine, Cosine,and Tangent respectively . Function cot Secant Note: The formulas provided are in respect to the picture. sec Cosecan cosec t Hypotenus e Adjacent Hypotenus e Opposite Adjacent Adjacent Opposite Hypotenus e Adjacent Hypotenus e THE TRIGONOMETRIC VALUES Angle A 0o 30o 45o 60o 90o sin A 0 1 1 √3 1 1 2 √3 √2 1 2 1 0 0 2 1 √2 1 2 √3 √3 2 √2 2 2 √2 √3 2 √3 √3 1 1 cos A tan A cosec A sec A cot A Not Defined 1 Not Defined √3 Not Defined 1 Not Defined 0 nometric your room !! Perpendicular = 6 mts. Hypotenuse = 12mts. = Sin 30 o Angle of Elevation = 30o 1. THANK YTrigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 2.  The word ‘trigonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure).  Trigonometry is the study of relationships between the sides and angles of a triangle.  Early astronomers used it to find out the distances of the stars and planets from the Earth.  Even today, most of the technologically advanced methods used in Engineering
  • 4. and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 3.  A triangle in which one angle is equal to 90 is called right triangle.  The side opposite to the right angle is known as hypotenuse. AC is the hypotenuse  The other two sides are known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights Reserved 4.  In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = (Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 5. Pythagoras Theorem Proof:  Given: Δ ABC is a right angled triangle where B = 900 And AB = P, BC= b and AC = h.  To Prove: h2 = p2 + b2  Construction : Draw a BD from B to AC , where AD = x and CB = h-x ,  Proof : In Δ ABC and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ ABC and Δ BDC both are similar So by these similarity, p b h A B C 6. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A B C 7.  Let us take a right triangle ABC  Here, ∠ ACB ( ) is an acute angle.  The position of the side AB with respect to angle . We call it the side opposite to angle .  AC is the hypotenuse of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTuto r. 2000-2013. All Rights Reserved 8.  The trigonometric ratios of the angle C in right ABC as follows :  Sine of C = =  Cosine of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 9.  Tangent of C = =  Cosecant of C= =  Secant of C = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved 10.  Cotangent of C  Above Trigonometric Ratio arbitrates as sin C, cos C, tan C , cosec C , sec C, Cot C .  If the measure of angle C is ‘ ’ then the ratios are : sin , cos , tan , cosec , sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 11.  Tan =  Cosec = 1 / Sin  Sec = 1 / Cos  Cot = Cos / Sin = 1 / Tan A CB p b h © iTutor. 2000- 2013. All Rights Reserved cos sin 12. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h © iTutor. 2000-2013. All Rights Reserved 13. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000 - 2013. All Rights Reserved 14.  Sin 450 = = = = 1/ 2  Cos 450 = = = = 1/ 2  Tan 450 = = = = 1  Cosec 450 = 1 / sin 450 = 1 / 1/ 2 = 2  Sec 450 = 1 / cos 450 = 1 / 1/ 2 = 2  Cot 450 = 1 / tan 450 = 1 / 1 = 1 Side opposite to 450
  • 5. Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved 15.  Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD ----- ---- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved 16.  As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 17. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric rat ios of 600 AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 18. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 19. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 20.  Relation of with Sin when 00 900 The greater the value of ‘ ’, the greater is the value of Sin . Smallest value of Sin = 0 Greatest value of Sin = 1  Relation of with Cos when 00 900 The greater the value of ‘ ’, the smaller is the value of Cos . Smallest value of Cos = 0 Greatest value of Cos = 1 © iTutor. 2000-2013. All Rights Reserved 21.  Relation of with tan when 00 900 Tan increases as ‘ ’ increases But ,tan is not defined at ‘ ’ = 900 Smallest value of tan = 0 © iTutor. 2000-2013. All Rights Reserved 22.  If 00 900 1. Sin(900- ) = Cos 2. Cos(900- ) = Sin  If 00< 900 1. Tan(900- ) = Cot 2. Sec(900- ) = Cosec  If 00 < 900 1. Cot(900- )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All Rights Reserved 23. Sin2 +Cos2 = 1 Sec2 -Tan2 = 1 Cosec2 - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved 24. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 25. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 26. relationships between the sides and ang technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 27.
  • 6. known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights Reserved 28. sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = (Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 29. rom B to AC , where AD = x and CB = h- --------(AA) In Δ ABC and Δ BDC both are similar So by these similarity, p b h A B C 30. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A B C 31. ∠ side AB with respect to angle . We call it the si of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved 32. of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 33. adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved 34. sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 35. 2000-2013. All Rights Reserved cos sin 36. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h © iTutor. 2000-2013. All Rights Reserved 37. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000-2013. All Rights Reserved 38. 450 = 1 / 1/ 2 Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved
  • 7. 39. therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved 40. he sides of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 41. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 42. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 43. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 44. -2013. All Rights Reserved 45. 2000-2013. All Rights Reserved 46. - ) = Cos 2. Cos(900- - ) = Cot 2. Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All Rights Reserved 47. - - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved 48. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 49. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 50. gonometry’ is derived from the Greek words ‘tri’(meaning three), ‘gon’ out the dist technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 51. known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights Reserved
  • 8. 52. sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 = (Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 53. , where AD = x and CB = h- C and Δ ABD, Δ ABC Δ ABD --------(AA) In Δ ABC and Δ BDC both are similar So by these similarity, p b h A B C 54. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A B C 55. ∠ of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved 56. of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 57. tetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved 58. in C, cos C, tan C , cosec C , sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 59. 2000-2013. All Rights Reserved cos sin 60. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h © iTutor. 2000-2013. All Rights Reserved 61. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000-2013. All Rights Reserved 62. tan 450 = 1 / 1 = 1 Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved 63. le in an equilateral triangle is 60°, therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT)
  • 9. Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved 64. of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 65. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 66. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 67. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 68. greater the value of ‘ ’, the greater is the value of e of Cos = 1 © iTutor. 2000-2013. All Rights Reserved 69. -2013. All Rights Reserved 70. - ) = Cos 2. Cos(900- - ) = Cot 2. Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All Rights Reserved 71. - - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved 72. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 73. Trigonometry Adjacent Opposite T- 1-855-694-8886 Email- info@iTutor.com By iTutor.com 74. ee), ‘gon’ most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts. © iTutor. 2000-2013. All Rights Reserved 75. site to known as legs. AB and BC are the legs Trigonometry deals with Right Triangles A CB © iTutor. 2000-2013. All Rights Reserved 76. any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of areas of the squares whose sides are the two legs. A CB (Hypotenuse)2 =
  • 10. (Perpendicular)2 + (Base)2 AC2 = BC2 + AB2 © iTutor. 2000-2013. All Rights Reserved 77. Pyt , where AD = x and CB = h- --------(AA) In Δ ABC and Δ BDC both are similar So by these similarity, p b h A B C 78. Or P2 = x × h And b2 = h (h – x) Adding both L.H.S. and R.H. S. Then p2 + b2 = (x × h) + h (h – x) Or p2 + b2 = xh + h2 – hx Hence the Pythagoras theorem p2 + b2 = h2 b xh h b p x h p And p b h A B C 79. ∠ of the right triangle and the side BC is a part of . So, we call it the side adjacent to angle . A CB Sideoppositetoangle Side adjacent to angle ‘ ’ © iTutor. 2000 -2013. All Rights Reserved 80. of C= = A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Hypotenuse AB AC Side adjacent to C Hypotenuse BC AC © iTutor. 2000-2013. All Rights Reserved 81. adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC Side adjacent to C Hypotenuse Side opposite to C Hypotenuse AC AB AC AB = © iTutor. 2000-2013. All Rights Reserved 82. sec sec and cot A CB Sideoppositetoangle Side adjacent to angle ‘ ’ Side opposite to C Side adjacent to C AB BC= = © iTutor. 2000-2013. All Rights Reserved 83. 2000-2013. All Rights Reserved cos sin 84. 1. Sin = p / h 2. Cos = b / h 3. Tan = p / b 4. Cosec = h / p 5. Sec = h / b 6. Cot = b / p A CB p b h © iTutor. 2000-2013. All Rights Reserved 85. Trigonometric Ratios of 45° In Δ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠ C = 45° So, BC = AB Now, Suppose BC = AB = a. Then by Pythagoras Theorem, AC2 = BC2 + AB2 = a2 + a2 AC2 = 2a2 , or AC = a 2 A CB 450 a a 450 © iTutor. 2000-2013. All Rights Reserved 86. Side opposite to 450 Hypotenuse AB AC a a 2 Side adjacent to 450 Hypotenuse BC AC a Side opposite to 450 Side adjacent to 450 AB BC a a a 2 © iTutor. 2000-2013. All Rights Reserved 87. °, therefore, ∠ A = ∠ B = ∠ C = 60°. Draw the perpendicular AD from A to the side BC, Now Δ ABD ≅ Δ ACD --------- (S. A. S) Therefore, BD = DC and ∠ BAD = ∠ CAD -----------(CPCT) Now observe that: Δ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60° 600 300 A B D C © iTutor. 2000-2013. All Rights Reserved
  • 11. 88. of the triangle. So, let us suppose that AB = 2a. BD = ½ BC = a AD2 = AB2 – BD2 = (2a)2 - (a)2 = 3a2 AD = a 3 Now Trigonometric ratios Sin 300 = = = = ½ 600 300 A B D C 2a 2a 2a a aSide opposite to 300 Hypotenuse BD AB a 2a © iTutor. 2000-2013. All Rights Reserved 89. Cos 300 = = = 3 / 2 Tan 300 = = = 1 / 3 Cosec 300 = 1 / sin 300 = 1 / ½ = 2 Sec 300 = 1 / cos 300 = 1 / 3/2 = 2 / 3 Cot 300 = 1 / tan 300 = 1 / 1/ 3 = 3 Now trigonometric ratios of 600 AD AB a 3 2a BD AD a a 3 300 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 90. Sin 600 = = = 3 / 2 Cos 600 = = = ½ Tan 600 = = = 3 Cosec 600 = 1 / Sin 600 = 1 / 3 / 2 = 2 / 3 Sec 600 = 1 / cos 600 = 1 / ½ = 2 Cot 600 = 1 / tan 600 = 1 / 3 AD AB a 3 2a BD AB a 2a AD BD a 3 a 600 A B D C 2a 2a 2a a a © iTutor. 2000-2013. All Rights Reserved 91. T. Ratios 0 30 45 60 90 Sine 0 ½ 1/ 2 3/2 1 Cosine 1 3/2 1/ 2 ½ 0 Tangent 0 1/ 3 1 3 Not defined Cosecant Not defined 2 2 2/ 3 1 Secant 1 2/ 3 2 2 Not defined Cotangent Not defined 3 1 1/ 3 0 © iTutor. 2000-2013. All Rights Reserved 92. ter is the value of -2013. All Rights Reserved 93. -2013. All Rights Reserved 94. - ) = Cos 2. Cos(900- 1. Tan(900- ) = Cot 2. Sec(900- - )= Tan 2. Cosec(900- ) = Sec A CB p b h © iTutor. 2000-2013. All Rights Reserved 95. - - Cot2 = 1 © iTutor. 2000-2013. All Rights Reserved 96. The End Call us for more information: www.iTutor.com 1-855-694-8886 Visit 97.