Question

i am trying to reduce this further… what do i do..
y^'= 2 sinx(cos(x)sin(2x)+sin(x)cos(2x))

Answer 1

\[y \prime= 2sinx(cosxsin2x+sinxcos2x)\]

Answer 2

The part with cos(x)sin(2x)+sin(x)cos(2x) looks like the angle addition identity for sine:
\[\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)\]
So, you can simplify that chunk to sin(x+2x), or sin(3x).

Answer 3

my final answer should be \[2\sin^2(x)(2\cos(2x)+1)\]

Answer 4

where do you get the sin^2?

Answer 5

I'd get: \[2 \sin (x) \sin(3x)\]

Answer 6

no idea. he gives us answers to the problem but we have to show steps and that is answer he has given

Answer 7

I'm not sure how to get there...sorry!

Answer 8

thanks anyways! i don't think he uses identities.