# Thread:

1. ## Double angle formulae

Can someone please show me the process of answering this question?

By writing sin3X as sin(2X+X), show that sin3X=3sinX−4sin^3X

I know that sin2X=2sinXcosX and the other trig identities, but I cannot understand how to substitute it in, in particular the answer follows as:

sin3X = sin2XcosX + cos2XsinX - it is this step that I am struggling with!

Thanks

2. Originally Posted by CSG18
Can someone please show me the process of answering this question?

By writing sin3X as sin(2X+X), show that sin3X=3sinX−4sin^3X

I know that sin2X=2sinXcosX and the other trig identities, but I cannot understand how to substitute it in, in particular the answer follows as:

sin3X = sin2XcosX + cos2XsinX - it is this step that I am struggling with!

Thanks
Use the hint given at the start of the question and recall that $\sin(A+B) = \sin A \cos B + \cos B \sin A$

$\sin(2X+X) = \sin (2x) \cos (x) + \cos (2x) \sin(x)$
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