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#