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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

Verified Answer :

MissPhiladelphia

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.

Answer:

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

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

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

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

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

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

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