How do you prove $sin (a + b) + sin (a + b) = 2 sin a cos b$ $sin(a+b) + sin(a+b) = 2sinacosb$ ?

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Answer 1 · 2016-03-27T01:11:17

You cannot prove it. It is not always true. But it is true that $sin (a + b) + sin (a - b) = 2 sin a cos b$ $sin(a+b)+sin(a-b) = 2sinacosb$

$sin (a + b) = sin a cos b + cos a sin b$ $sin(a+b) = sinacosb+cosasinb$ .

So,

$sin (a + b) + sin (a + b) = 2 sin (a + b)$ $sin(a+b)+sin(a+b) = 2sin(a+b)$

$= 2 (sin a cos b + cos a sin b)$ $= 2(sinacosb+cosasinb)$

$= 2 sin a cos b + 2 cos a sin b$ $= 2sinacosb+2cosasinb$ .

The last is equal to $2 sin a cos b$ $2sinacosb$ only if $cos a = 0$ $cosa = 0$ or $sin b = 0$ $sinb = 0$

But

$sin (a - b) = sin a cos b - cos a sin b$ $sin(a-b) = sinacosb-cosasinb$ .

So,

$sin (a + b) + sin (a - b) = (sin a cos b + cos a sin b) + (sin a cos b - cos a sin b)$ $sin(a+b)+sin(a-b) = (sinacosb+cosasinb)+(sinacosb-cosasinb)$

$= 2 sin a cos b$ $= 2sinacosb$

How do you prove sin(a+b) + sin(a+b) = 2sinacosb sin(a+b)+sin(a+b)=2sinacosb sin(a+b) + sin(a+b) = 2sinacosb ?