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

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

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

So,

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

$= 2(sinacosb+cosasinb)$

$= 2sinacosb+2cosasinb$ .

The last is equal to $2sinacosb$ only if $cosa = 0$ or $sinb = 0$

But

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

So,

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

$= 2sinacosb$

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