Trigonometry Question I tried but can't seem to figure out the answer (1 Viewer)

[Leffife]

Answer: (1/4 sin 4x)

Simplify:

sin x cos x cos 2x

My working:

= sinx cosx (1 - 2sin²x)
= sinx cosx - 2sin³x cosx
= cosx(sinx - 2sin³x)
= sinxcosx (1 - 2sin²x)
= sinxcosx (1 - 2(1-cos²x))
= sinxcosx (1 - 2 + cos²x)
= sinxcosx (cos²x - 1)
= sinxcosx (-sin²x)
= -sin³xcosx

I'm just stuck now.

 

[nightweaver066]

Recognise that sin2x = 2sinxcosx

So sincosx = sin2x/2



Similarly, sin4x = 2sin2xcos2x, do the same thing and you get your answer.

 

[irenefu]

= 1/2sin2xcos2x
=1/2 x 1/2 sin4x
= 1/4 sin 4x

I think that should be right

 

[bleakarcher]

sin(x)cos(x)cos(2x)
=(1/2)[2sin(x)cos(x)]cos(2x)
=(1/2)sin(2x)cos(2x)
=(1/2)(1/2)[2sin(2x)cos(2x)
=(1/4)sin(4x)