Question

Rewrite with only sin x and cos x.

sin 3x
a)2 sin x cos2x + cos x
b)2 sin x cos2x + sin3x
c)sin x cos2x - sin3x + cos3x
d)2 cos2x sin x + sin x - 2 sin3x

Answer 1

d) cos2x sin x + 2sin x - 2 sin^3x

Step-by-step explanation:

As we know that

`sin(3x) = sin(2x + x)`

now we will have

`sin(3x) = sin2xcosx + cos2xsinx`

now we also know that

`sin(2x) = 2sinx cosx`

now we can use this in above equation

`sin(3x) = (2sinxcosx)cosx + cos(2x)sinx`

`sin(3x) = 2sinx(cos^2x) + cos(2x)sinx`

`sin(3x) = 2sinx(1 - sin^2x) + cos(2x)sinx`

`sin(3x) = 2sinx - 2sin^3x + cos(2x)sinx`

Answer 2

For the answer to the question above,
I'll write down my solution to your problem
sin(A+B) = sinAcosB + cosAsinB
sin(2A) = 2sinAcosA
cos(2A) = 1-2sin^2A
sin(3x) = sin(2x+x)

sin(3x) = sin(2x)cos(x) + cos(2x)sin(x)
= 2sin(x)cos(x)cos(x) + (1-2sin^2(x))sin(x)
= 2sin(x)cos^2(x) + sin(x) - 2sin^3(x)
= 2sin(x)(1-sin^2(x)) + sin(x) - 2sin^3(x)
= 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)
= 3sin(x) - 4sin^3(x)
My closest answer is multiple choice letter D.