# Help me prove: sin(A+B) = sinA cosB + cosA sinB [duplicate]

Can you help me prove that: sin(A+B) = sinA cosB + cosA sinB?

Thanks!

-

## marked as duplicate by littleO, 1999, amWhy, studiosus, Rebecca J. StonesJul 9 '14 at 1:21

See here. –  Lucian Jul 8 '14 at 23:30
This answer, in particular. –  Omnomnomnom Jul 8 '14 at 23:41
This question also has several helpful answers –  Omnomnomnom Jul 8 '14 at 23:43

Here are two for the price of one (using Euler's formula):

$$\color{red}{\cos(A+B)}+i\color{green}{\sin(A+B)} \equiv e^{i(A+B)} \equiv e^{iA} \times e^{iB}$$ $$\equiv [\cos(A)+i\sin(A)][\cos(B)+i\sin(B)]$$ $$\equiv \color{red}{[\cos(A)\cos(B)-\sin(A)\sin(B)]}+i\color{green}{[\sin(A)\cos(B)+\cos(A)\sin(B)]}$$

Now equate imaginary parts to give the result for $\sin(A+B)$ (and, if you want, equate real parts to give the result for $\cos(A+B)$).

-
Isnt this circular ? How for example do you prove $e^{ix}=\cos x +i\sin x$? –  Rene Schipperus Jul 8 '14 at 23:40
@ReneSchipperus How is it circular? The fact that $e^{ix} \equiv \cos(x)+i \sin(x)$ does not rely on the statement posed by the OP. –  alexqwx Jul 8 '14 at 23:42
Oh I guess you do that from the power series definition –  Rene Schipperus Jul 8 '14 at 23:42
@ReneSchipperus Yep, and there are many more proofs! See en.wikipedia.org/wiki/Euler's_formula#Proofs . –  alexqwx Jul 8 '14 at 23:43
There are some derivations of Eulers formula which rely on the sum formula. –  Rene Schipperus Jul 8 '14 at 23:43

Hint: write it as $\sin(x) = 1/2 i e^{-i x} - 1/2 i e^{i x}$

-
Hey, sorry I'm still pretty new to math; do you think you can write out your proof all the way? I think I get where you're going, but I'm not totally sure. :) Thanks! –  user162370 Jul 9 '14 at 1:17

Here is a proof without using complex numbers, from Apostol.

$$\sin(x+y) = -\cos(x + y + \frac{\pi}{2}) = -\cos x\cos(y + \frac{\pi}{2}) + \sin(x)\ \sin(y + \frac{\pi}{2})$$

$$= \cos x\ \sin y + \sin x\cos y$$

Assuming you can prove the formula for the cosine, of course, which Apostol gives as a property of the cosine.

-

I found a very interesting geometric proof here. It proves it relatively clearly and concisely.

-

May be a little unconventional, but assuming we have an isosceles right angled triangle.

Then, $sin 90^\circ=1$

$$=\frac{1}{2}+\frac{1}{2}$$

$$=\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}$$

$$=cos 45^\circ \cdot sin 45^\circ+sin 45^\circ \cdot cos 45^\circ$$

The similar can be proved for a scalene triangle as well.

-
This doesn't really prove anything in general, though.... –  David Z Jul 9 '14 at 0:31