# How can you prove, sin3x = sinx(3cos^2x - sin^2x). Is there an identity that can be used?

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You can first use the law of sine addition first:

Where "a" is 2x and "b" is x. So, we would have:

sin(2x+x) = sin2x cosx + sinx cos2x

Then, we use the double angle formulas for sin and cos:

Plugging these into what we have so far:

sin(2x+x) = sin2x cosx + sinx cos2x

= 2 sinx cosx cosx + sinx (cos^2x - sin^2x)

= 2 sinx cos^2x + sinx cos^2x - sin^3x

= 3 sinx cos^2x - sin^3x

= sinx (3 cox^2x - sin^2x)

using an identity of sin (3x) = 3sin (X) - 4 sin ^3 (x)

sin (3x) = 3sin(x) - 4sin^3(x)

= sin(x) [ 3*(1)-4 sin^2 (x) ]

here sin^2 (x) + cos^2 (x) =1 identity is used

= sin(x)[ 3*[sin^2 (x) + cos^2 (x) ] - 4 sin^2 (x) ]

= sin(x)( 3cos^2 (x) ) - sin ^2 (x) )

I hope this will help you ......:)

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