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Zorluk: ZorFundamental Trigonometric Identities

An angle θ\theta lies in the third quadrant, where π<θ<3π2\pi < \theta < \frac{3\pi}{2}. If sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5}, what is the value of the expression 125(sin⁡3θ+cos⁡3θ)125(\sin^3\theta + \cos^3\theta)?

Cevap: -91

Cevap

The correct value of the expression is -91.
Squaring the equation sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5} gives 1−2sin⁡θcos⁡θ=1251 - 2\sin\theta\cos\theta = \frac{1}{25}, which simplifies to sin⁡θcos⁡θ=1225\sin\theta\cos\theta = \frac{12}{25}. We then find the square of the sum: (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ=1+2425=4925(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta = 1 + \frac{24}{25} = \frac{49}{25}. Since the angle is in the third quadrant, both trigonometric functions are negative, so we choose the negative root sin⁡θ+cos⁡θ=−75\sin\theta + \cos\theta = -\frac{7}{5}. Factoring the sum of cubes gives sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(sin⁡2θ−sin⁡θcos⁡θ+cos⁡2θ)=(−75)(1−1225)=−91125\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (-\frac{7}{5})(1 - \frac{12}{25}) = -\frac{91}{125}. Multiplying this by 125 yields -91.

Adım Adım Çözüm

1
Square both sides of the equation sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5}
sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=125  ⟹  1−2sin⁡θcos⁡θ=125  ⟹  sin⁡θcos⁡θ=1225\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = \frac{1}{25} \implies 1 - 2\sin\theta\cos\theta = \frac{1}{25} \implies \sin\theta\cos\theta = \frac{12}{25}
To solve for the product of sine and cosine using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
2
Determine the value of sin⁡θ+cos⁡θ\sin\theta + \cos\theta using the identity (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta
(sin⁡θ+cos⁡θ)2=1+2(1225)=4925  ⟹  sin⁡θ+cos⁡θ=−75(\sin\theta + \cos\theta)^2 = 1 + 2(\frac{12}{25}) = \frac{49}{25} \implies \sin\theta + \cos\theta = -\frac{7}{5}
Because θ\theta is in the third quadrant (π<θ<3π2\pi < \theta < \frac{3\pi}{2}), both sin⁡θ\sin\theta and cos⁡θ\cos\theta must be negative, meaning their sum is also negative.
3
Use the sum of cubes factorization to evaluate sin⁡3θ+cos⁡3θ\sin^3\theta + \cos^3\theta
sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(sin⁡2θ−sin⁡θcos⁡θ+cos⁡2θ)=(−75)(1−1225)=−91125\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (-\frac{7}{5})(1 - \frac{12}{25}) = -\frac{91}{125}
To express the sum of cubes in terms of the known sum and product of sine and cosine.
4
Multiply the evaluated sum of cubes by 125
125×(−91125)=−91125 \times (-\frac{91}{125}) = -91
To compute the final value of the requested expression.

Anahtar Kavram

Pythagorean trigonometric identities, quadrant sign analysis, and algebraic factorization of the sum of cubes

Alternatif Yöntem

Instead of applying algebraic identities to find the sum of cubes directly, we can solve for the individual values of sin⁡θ\sin\theta and cos⁡θ\cos\theta from the system of equations: sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5} and sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Substituting sin⁡θ=cos⁡θ−15\sin\theta = \cos\theta - \frac{1}{5} into the second equation yields 2cos⁡2θ−25cos⁡θ−2425=02\cos^2\theta - \frac{2}{5}\cos\theta - \frac{24}{25} = 0, which factors as (5cos⁡θ+3)(5cos⁡θ−4)=0(5\cos\theta + 3)(5\cos\theta - 4) = 0. Since θ\theta is in Quadrant III, cos⁡θ=−35\cos\theta = -\frac{3}{5} and sin⁡θ=−45\sin\theta = -\frac{4}{5}. Evaluating 125(sin⁡3θ+cos⁡3θ)125(\sin^3\theta + \cos^3\theta) directly with these values gives 125((−45)3+(−35)3)=125(−64125−27125)=−91125((-\frac{4}{5})^3 + (-\frac{3}{5})^3) = 125(-\frac{64}{125} - \frac{27}{125}) = -91.
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Fundamental Trigonometric Identities

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