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2 July, 05:00

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

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Answers (1)
  1. 2 July, 05:18
    0
    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.
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