How do you express sin 3thetasin3θ in terms of trigonometric functions of thetaθ ?
2 Answers
Explanation:
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We have double-angle identities:
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+3icos2θsinθ−3cosθsin2θ−isin3θ
= (cos^3 theta - 3 cos theta sin^2 theta) + i (3 cos^2 theta sin theta - sin^3 theta)=(cos3θ−3cosθsin2θ)+i(3cos2θsinθ−sin3θ)
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