Help me prove: sin(A+B) = sinA cosB + cosA sinB

Question

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

Thanks!

This answer, in particular. – Ben Grossmann Jul 08 '14 at 23:41 — Ben Grossmann, Jul 08 '14 at 23:41
This question also has several helpful answers – Ben Grossmann Jul 08 '14 at 23:43 — Ben Grossmann, Jul 08 '14 at 23:43

score 13 · Accepted Answer · answered Jul 08 '14 at 23:35

13

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 $\cos(A+B)$).

answered Jul 08 '14 at 23:35

beep-boop

11,595

3

Isnt this circular ? How for example do you prove $e^{ix}=\cos x +i\sin x$? – Rene Schipperus Jul 08 '14 at 23:40
1

@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. – beep-boop Jul 08 '14 at 23:42
Oh I guess you do that from the power series definition – Rene Schipperus Jul 08 '14 at 23:42
@ReneSchipperus Yep, and there are many more proofs! See http://en.wikipedia.org/wiki/Euler's_formula#Proofs . – beep-boop Jul 08 '14 at 23:43
There are some derivations of Eulers formula which rely on the sum formula. – Rene Schipperus Jul 08 '14 at 23:43
@ReneSchipperus Ah- I didn't know that. Do you have a link to such a proof? – beep-boop Jul 08 '14 at 23:44
No link, but if you use DeMoivre's formula and the limit definition of $e^x$ you can see how its done. – Rene Schipperus Jul 08 '14 at 23:46

score 2 · Answer 2 · answered Jul 08 '14 at 23:30

2

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

answered Jul 08 '14 at 23:30

Loreno Heer

4,460

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! – Gauss Jul 09 '14 at 01:17

score 2 · Answer 3 · answered Jul 09 '14 at 01:07

2

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

answered Jul 09 '14 at 01:07

Gauss

395

score 1 · Answer 4 · edited Jul 09 '14 at 00:15

1

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.

edited Jul 09 '14 at 00:15

Ivo Terek

77,665

answered Jul 08 '14 at 23:55

Lucas Virgili

183

score -3 · Answer 5 · answered Jul 08 '14 at 23:52

-3

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.

answered Jul 08 '14 at 23:52

MonK

1,794

This doesn't really prove anything in general, though.... – David Z Jul 09 '14 at 00:31

Help me prove: sin(A+B) = sinA cosB + cosA sinB

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