Application of Euler\'s formula: Prove the following identities.
1.cos(A - B) = cosAcosB + sinAsinB and
sin(A - B) = sinAcosB - cosAsinB
(Hint: The cosine and sine functions are even and odd, respectively.)
2.State a single expression for both cos(A ± B) and sin(A ± B)
Solution
Let cis(N) represents cos(N) + i*sin(N).
1. Start with cis(A-B)
=e^(A-B)i, by Euler\'s formula.
=e^(Ai) * e^(-Bi)
=cis(A) * cis(-B)
=(cos(A)+isin(A))*(cos(-B)+isin(-B))
=cos(A)cos(-B) + isin(A)*cos(-B) + isin(-B)*cos(A) + i^2sin(A)sin(-B)
Remember that i^2=-1
=cos(A)cos(-B) + isin(A)*cos(-B) + isin(-B)*cos(A) - sin(A)sin(-B)
=cos(A)cos(-B)-sin(A)sin(-B) + i[sin(A)cos(-B)+cos(A)sin(-B)]
Now, recall that cos(-B)=cos(B), and sin(-B)=-sin(B):
=cosAcosB+sinAsinB + i[sinAcosB-cosAsinB]
Which we recall is equal to cis(A-B) = cos(A-B) + i sin(A-B).
Matching the real parts of each side gives:
cos(A-B) = cosAcosB+sinAsinB.
And matching the imaginary parts of each side gives:
sin(A-B) = sinAcosB-cosAsinB
2. Doing the same for cis(A+B) gives:
cos(A+B) = cosAcosB-sinAsinB
sin(A+B) = sinAcosB+cosAsinB
So, putting parts 1 and 2 together gives: cos(A±B) = cosAcosB(-/+)sinAsinB
and
sin(A±B) = sinAcosB±cosAsinB.
In the cos formula, (-/+) indicates that you subtract for A+B, and add for A-B..
Application of Eulers formula Prove the following identities. 1.pdf
1. Application of Euler's formula: Prove the following identities.
1.cos(A - B) = cosAcosB + sinAsinB and
sin(A - B) = sinAcosB - cosAsinB
(Hint: The cosine and sine functions are even and odd, respectively.)
2.State a single expression for both cos(A ± B) and sin(A ± B)
Solution
Let cis(N) represents cos(N) + i*sin(N).
1. Start with cis(A-B)
=e^(A-B)i, by Euler's formula.
=e^(Ai) * e^(-Bi)
=cis(A) * cis(-B)
=(cos(A)+isin(A))*(cos(-B)+isin(-B))
=cos(A)cos(-B) + isin(A)*cos(-B) + isin(-B)*cos(A) + i^2sin(A)sin(-B)
Remember that i^2=-1
=cos(A)cos(-B) + isin(A)*cos(-B) + isin(-B)*cos(A) - sin(A)sin(-B)
=cos(A)cos(-B)-sin(A)sin(-B) + i[sin(A)cos(-B)+cos(A)sin(-B)]
Now, recall that cos(-B)=cos(B), and sin(-B)=-sin(B):
=cosAcosB+sinAsinB + i[sinAcosB-cosAsinB]
Which we recall is equal to cis(A-B) = cos(A-B) + i sin(A-B).
Matching the real parts of each side gives:
cos(A-B) = cosAcosB+sinAsinB.
And matching the imaginary parts of each side gives:
sin(A-B) = sinAcosB-cosAsinB
2. Doing the same for cis(A+B) gives:
cos(A+B) = cosAcosB-sinAsinB
sin(A+B) = sinAcosB+cosAsinB
So, putting parts 1 and 2 together gives: cos(A±B) = cosAcosB(-/+)sinAsinB
and
sin(A±B) = sinAcosB±cosAsinB.
In the cos formula, (-/+) indicates that you subtract for A+B, and add for A-B.
![Application of Euler's formula: Prove the following identities.
1.cos(A - B) = cosAcosB + sinAsinB and
sin(A - B) = sinAcosB - cosAsinB
(Hint: The cosine and sine functions are even and odd, respectively.)
2.State a single expression for both cos(A ± B) and sin(A ± B)
Solution
Let cis(N) represents cos(N) + i*sin(N).
1. Start with cis(A-B)
=e^(A-B)i, by Euler's formula.
=e^(Ai) * e^(-Bi)
=cis(A) * cis(-B)
=(cos(A)+isin(A))*(cos(-B)+isin(-B))
=cos(A)cos(-B) + isin(A)*cos(-B) + isin(-B)*cos(A) + i^2sin(A)sin(-B)
Remember that i^2=-1
=cos(A)cos(-B) + isin(A)*cos(-B) + isin(-B)*cos(A) - sin(A)sin(-B)
=cos(A)cos(-B)-sin(A)sin(-B) + i[sin(A)cos(-B)+cos(A)sin(-B)]
Now, recall that cos(-B)=cos(B), and sin(-B)=-sin(B):
=cosAcosB+sinAsinB + i[sinAcosB-cosAsinB]
Which we recall is equal to cis(A-B) = cos(A-B) + i sin(A-B).
Matching the real parts of each side gives:
cos(A-B) = cosAcosB+sinAsinB.
And matching the imaginary parts of each side gives:
sin(A-B) = sinAcosB-cosAsinB
2. Doing the same for cis(A+B) gives:
cos(A+B) = cosAcosB-sinAsinB
sin(A+B) = sinAcosB+cosAsinB
So, putting parts 1 and 2 together gives: cos(A±B) = cosAcosB(-/+)sinAsinB
and
sin(A±B) = sinAcosB±cosAsinB.
In the cos formula, (-/+) indicates that you subtract for A+B, and add for A-B.](https://image.slidesharecdn.com/applicationofeulersformulaprovethefollowingidentities-230325094239-910da0bc/85/Application-of-Eulers-formula-Prove-the-following-identities-1-pdf-1-320.jpg)
