How do you express sin 3thetasin3θ in terms of trigonometric functions of thetaθ ?

2 Answers
Jan 4, 2018

sin3x=3sinxcos^2x-sin^3xsin3x=3sinxcos2xsin3x

Explanation:

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sin3x=sin(2x+x)=sin2xcosx+cos2xsinxsin3x=sin(2x+x)=sin2xcosx+cos2xsinx

We have double-angle identities:

sin2x=2sinxcosxsin2x=2sinxcosx

cos2x=cos^2x-sin^2xcos2x=cos2xsin2x

sin3x=2sinxcosxcosx+(cos^2x-sin^2x)sinxsin3x=2sinxcosxcosx+(cos2xsin2x)sinx

sin3x=2sinxcos^2x+sinxcos^2x-sin^3xsin3x=2sinxcos2x+sinxcos2xsin3x

sin3x=3sinxcos^2x-sin^3xsin3x=3sinxcos2xsin3x

Jan 4, 2018

sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta = 3 sin theta - 4 sin^3 thetasin3θ=3cos2θsinθsin3θ=3sinθ4sin3θ

Explanation:

We can use de Moivre's formula:

(cos theta + i sin theta)^n = cos n theta + i sin n theta(cosθ+isinθ)n=cosnθ+isinnθ

as follows:

cos 3 theta + i sin 3 thetacos3θ+isin3θ

= (cos theta + i sin theta)^3=(cosθ+isinθ)3

= cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin^2 theta - i sin^3 theta=cos3θ+3icos2θsinθ3cosθsin2θisin3θ

= (cos^3 theta - 3 cos theta sin^2 theta) + i (3 cos^2 theta sin theta - sin^3 theta)=(cos3θ3cosθsin2θ)+i(3cos2θsinθsin3θ)

Then equating real and imaginary parts, we find:

{ (cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta), (sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta) :}

We can also use the trigonometric version of Pythagoras' theorem:

cos^2 theta + sin^2 theta = 1

to refactor these formulae as follows:

cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta

color(white)(cos 3 theta) = cos^3 theta - 3 cos theta (1-cos^2 theta)

color(white)(cos 3 theta) = 4cos^3 theta - 3 cos theta

sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta

color(white)(sin 3 theta) = 3 (1-sin^2 theta) sin theta - sin^3 theta

color(white)(sin 3 theta) = 3 sin theta - 4 sin^3 theta