Roots, sin θ \sin \theta and cos θ \cos \theta

Geometry Level 3

For a constant k k , the two roots of the quadratic equation 3 x 2 x + k = 0 3x^2-x+k=0 are sin θ \sin \theta and cos θ . \cos \theta. What is the value of 54 ( sin 3 θ + cos 3 θ ) ? 54(\sin^3 \theta+\cos^3 \theta)?

28 29 26 27

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1 solution

Chew-Seong Cheong
Sep 26, 2018

Relevant wiki: Vieta's Formula Problem Solving - Intermediate

If sin θ \sin \theta and cos θ \cos \theta are roots of 3 x 2 x + k = 0 3x^2 - x + k = 0 , then by Vieta's formula:

{ sin θ + cos θ = 1 3 . . . ( 1 ) sin θ cos θ = k 3 . . . ( 2 ) \begin{cases} \sin \theta + \cos \theta = \dfrac 13 & ...(1) \\ \sin \theta \cos \theta = \dfrac k3 & ...(2) \end{cases}

And we have:

( sin θ + cos θ ) 2 = sin 2 θ + 2 sin θ cos θ + cos 2 θ 1 9 = 1 + 2 k 3 k = 4 3 \begin{aligned} (\sin \theta + \cos \theta)^2 & = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta \\ \frac 19 & = 1 + \frac {2k}3 \\ \implies k & = - \frac 43 \end{aligned}

Also:

sin 3 θ + cos 3 θ = ( sin θ + cos θ ) ( sin 2 θ + cos 2 θ sin θ cos θ ) = 1 3 ( 1 + 4 9 ) = 13 27 \begin{aligned} \sin^3 \theta + \cos^3 \theta & = (\sin \theta + \cos \theta)(\sin^2 \theta + \cos^2 \theta - \sin \theta \cos \theta) \\ & = \frac 13 \left(1 + \frac 49\right) = \frac {13}{27} \end{aligned}

54 ( sin 3 θ + cos 3 θ ) = 26 \implies 54(\sin^3 \theta + \cos^3 \theta) = \boxed {26} .

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