If \sin\theta+\cos\theta=\sqrt2, what is the value of \sin^6\theta+\cos^6\theta?

Trigonometry Practice Test 1

Question : 22 of 40

Marks: +1, -0

\sinθ+\cosθ=√2

, what is the value of

\sin^6θ+\cos^6θ?

Solution:

\sinθ+\cosθ=√2

⇒(\sinθ+\cosθ)^2=2

⇒1+2\sinθ\cosθ=2

⇒\sinθ\cosθ=1/2

∴\sin^6θ+\cos^6θ

=(\sin^2θ)^3+(\cos^2θ)^3

=(\sin^2θ+\cos^2θ)(\sin^4θ+\cos^2θ-\sin^2θ\cos^2θ)

=[(\sin^2θ+\cos^2θ)^2-3\sin^2θ \cos^2θ]

=[1-3\sin^2θ \cos^2θ]

=1-3/4=1/4

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