To express: The trigonometric functions cos 3 θ and sin 3 θ in terms of cos θ and sin θ by the use of De Moivre’s Theorem with n = 3 .

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter I, Problem 47E
To determine

To express: The trigonometric functions cos3θ and sin3θ in terms of cosθ and sinθ by the use of De Moivre’s Theorem with n=3.

Expert Solution

The trigonometric functions cos3θ expressed as cos3θ=cos3θ3sin2θcosθ.

The trigonometric functions sin3θ expressed as sin3θ=3sinθcos2θsin3θ.

Explanation of Solution

Theorem used:

De Moivre’s Theorem:

If z=r(cosθ+isinθ) and n be a positive integer then,

zn=[r(cosθ+isinθ)]n=rn(cosnθ+isinnθ).

Calculation:

As when n=3 and r=1 substitute in De Moivre’s Theorem,

[1(cosθ+isinθ)]3=13(cos3θ+isin3θ)(cosθ+isinθ)3=(cos3θ+isin3θ)

Simplify the expression (cosθ+isinθ)3=(cos3θ+isin3θ) and obtain the trigonometric functions cos3θ and sin3θ in terms of cosθ and sinθ.

(cosθ+isinθ)3=(cos3θ+isin3θ)cos3θ+3cos2θ(isinθ)+3cosθ(isinθ)2+(isinθ)3=(cos3θ+isin3θ)cos3θ+(3cos2θsinθ)i3cosθsin2θisin3θ=(cos3θ+isin3θ)[i2=1andi3=i](cos3θ3sin2θcosθ)+i(3cos2θsinθsin3θ)=(cos3θ+isin3θ)

Equating real part and imaginary part of the above equation.

Thus, cos3θ=cos3θ3sin2θcosθ and sin3θ=3sinθcos2θsin3θ.

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