Verify the following trig identity: sin(3θ)−sinθ=2cos(2θ)sinθ
Here is my work so far.
sin(3θ)−sinθ=2cos(2θ)sinθ
LHS:sin(θ+2θ)−sinθ
sinθcos(2θ)+sin(2θ)cosθ−sinθ
sinθcos(2θ)+(2sinθcosθ)cosθ−sinθ
Where do I go from here? I think I should leave the first term in the line above as is, and try and manipulate the second two terms to equal sinθcos(2θ), then the LHS will add together to equal 2cos(2θ)sinθ, and the identity will be verified. How do you suggest I get there?
Any hints or advice would be appreciated.
EDIT:
I ended up verifying this identity using the identity frank000 mentioned in comments. Thanks to everyone for the input, it was all very helpful.
Answer
hint: also use sin(θ)=sin(2θ−θ)=sin(2θ)cos(θ)−sin(θ)cos(2θ)
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