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Is it ever possible that sin (A+B)= sin⁡ A+sin⁡ B? If so, explain how. If not, explain why not.

I really don't understand the sum identity for sine. It's frustrating me to the core. Somebody please help.

2 Answers

2 votes

The formula for the sine sum is


\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)

So, equation


\sin(a)\cos(b)+\cos(a)\sin(b) = \sin(a)+\sin(b)

Cannot be solved with standard ways. However, we can build special cases to make it work.

The most trivial example would be A=B=0. In this case, we have


\sin(0+0)=0 \sin(0) = 0,\quad\sin(0)+\sin(0)=0+0=0

If we work on this idea a little bit, we can produce examples like


A=2k_1\pi,\quad B=2k_2\pi

And again we have


\sin(A+B)=\sin(2\pi(k_1+k_2))=0


\sin(A)+\sin(B)=\sin(2k_1\pi)+\sin(2k_2\pi)=0+0=0

So, yes, it is possible that
\sin(A+B)=\sin(A)+\sin(B), although solving the equation directly would be quite difficult.

User Milan Gajjar
by
8.1k points
4 votes

Answer: yes, when A & B are on the quadrantals (0°, 90°, 180°, 270°) which results in a sum of 0

Explanation:

sin (A + B) = cos A · sin B + cos B · sin A

If cos A = 1 and cos B = 1, then you will end up with sin A + sin B.

The only angle where cos is equal to 1 is at 0°. Notice that sin is equal to 0.

It follows that when cos is equal to -1, sin is also 0. This occurs at 180°

The other option is when cos is equal to 0. This occurs at 90° and 270°.

All of these options result in sin A + sin B = 0

So, the only possibility when sin (A + B) = sin A + sin B is when it equals zero.

User Robin Zimmermann
by
7.8k points
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