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# C3 Trig Identity Help Watch

1. Hey, I'm having trouble proving that
2cos3x - sin2xsinx = 2cos2x cosx

I can get the left hand side down to
= 2cos3x - (2sinxcosx)sinx
= 2cos3x - 2sin2xsinxcosx

And then the right hand side down to
= 2(cos2x-sin2x)cosx
= (2cos2x-2sin2x)cosx
= 2cos3x-2sin2xcosx

From here I'm stumped as despite being identical bar the extra sinx I can't figure out if it's there by mistake or if I've made a mistake higher up in my working, any help?
2. One too many sinx's on your second line of working.

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3. (Original post by Krollo)
One too many sinx's on your second line of working.

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Thanks!

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