Is cos(3θ)=cos^3(θ)-3sin^2(θ)cos(θ) ?

The answer to whether cos(3θ)=cos^3(θ)-3sin^2(θ)cos(θ) is True

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prove cos(3θ)=cos^3(θ)-3sin^2(θ)cos(θ)

prove

cos (3 θ) = cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

True

Solution steps

cos (3 θ) = cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

Manipulating left side

cos (3 θ)

Rewrite using trig identities

= cos (θ) cos (2 θ) - 2 sin^{2} (θ) cos (θ)

Use the Double Angle identity:

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

= (cos^{2} (θ) - sin^{2} (θ)) cos (θ) - 2 sin^{2} (θ) cos (θ)

Expand

(cos^{2} (θ) - sin^{2} (θ)) cos (θ) - 2 sin^{2} (θ) cos (θ) : cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

= cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

(sin (x) + cos (x))^{2} = 1 + sin^{2} (x)

cos^{2} (B) - sin^{2} (B) = 2 cos^{2} (B) - 1

\frac{1}{1 + sin ( x )} = \frac{1 - sin ( x )}{cos ^{2} ( x )}

csc^{2} (x) cos^{2} (x) = \frac{1}{tan ^{2} ( x )}

cot (t) tan (t) = 1

Is cos(3θ)=cos^3(θ)-3sin^2(θ)cos(θ) ?
The answer to whether cos(3θ)=cos^3(θ)-3sin^2(θ)cos(θ) is True