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

Thanks!

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5 Answers

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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 $\cos(A+B)$).

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

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

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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.

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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.

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