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Difficulty: Very hardTrigonometric Ratios and Identities

For an acute angle θ\theta, the sum of its sine and cosine is 72\frac{\sqrt{7}}{2}. What is the value of sin⁡3(θ)+cos⁡3(θ)\sin^3(\theta) + \cos^3(\theta)?

  1. A
    574\frac{5\sqrt{7}}{4}
  2. B
    11716\frac{11\sqrt{7}}{16}
  3. 5716\frac{5\sqrt{7}}{16}Answer
  4. D
    778\frac{7\sqrt{7}}{8}

Answer

The correct answer is 5716\frac{5\sqrt{7}}{16}.
The correct answer is 5716\frac{5\sqrt{7}}{16}. Squaring the given equation sin⁡(θ)+cos⁡(θ)=72\sin(\theta) + \cos(\theta) = \frac{\sqrt{7}}{2} gives sin⁡2(θ)+2sin⁡(θ)cos⁡(θ)+cos⁡2(θ)=74\sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = \frac{7}{4}. Substituting the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 gives 1+2sin⁡(θ)cos⁡(θ)=741 + 2\sin(\theta)\cos(\theta) = \frac{7}{4}, which simplifies to sin⁡(θ)cos⁡(θ)=38\sin(\theta)\cos(\theta) = \frac{3}{8}. Factoring the sum of cubes yields sin⁡3(θ)+cos⁡3(θ)=(sin⁡(θ)+cos⁡(θ))(1−sin⁡(θ)cos⁡(θ))\sin^3(\theta) + \cos^3(\theta) = (\sin(\theta) + \cos(\theta))(1 - \sin(\theta)\cos(\theta)). Substituting the values results in (72)(1−38)=5716\left(\frac{\sqrt{7}}{2}\right)\left(1 - \frac{3}{8}\right) = \frac{5\sqrt{7}}{16}.

Step-by-Step Solution

1
Square both sides of the given equation sin⁡(θ)+cos⁡(θ)=72\sin(\theta) + \cos(\theta) = \frac{\sqrt{7}}{2}.
(sin⁡(θ)+cos⁡(θ))2=sin⁡2(θ)+2sin⁡(θ)cos⁡(θ)+cos⁡2(θ)=74(\sin(\theta) + \cos(\theta))^2 = \sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = \frac{7}{4}.
This sets up the expression to apply the Pythagorean trigonometric identity.
2
Substitute the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and solve for the product term.
1+2sin⁡(θ)cos⁡(θ)=74  ⟹  2sin⁡(θ)cos⁡(θ)=34  ⟹  sin⁡(θ)cos⁡(θ)=381 + 2\sin(\theta)\cos(\theta) = \frac{7}{4} \implies 2\sin(\theta)\cos(\theta) = \frac{3}{4} \implies \sin(\theta)\cos(\theta) = \frac{3}{8}.
This determines the value of the product of the sine and cosine ratios.
3
Apply the sum of cubes factorization formula to sin⁡3(θ)+cos⁡3(θ)\sin^3(\theta) + \cos^3(\theta).
sin⁡3(θ)+cos⁡3(θ)=(sin⁡(θ)+cos⁡(θ))(sin⁡2(θ)−sin⁡(θ)cos⁡(θ)+cos⁡2(θ))=(sin⁡(θ)+cos⁡(θ))(1−sin⁡(θ)cos⁡(θ))\sin^3(\theta) + \cos^3(\theta) = (\sin(\theta) + \cos(\theta))(\sin^2(\theta) - \sin(\theta)\cos(\theta) + \cos^2(\theta)) = (\sin(\theta) + \cos(\theta))(1 - \sin(\theta)\cos(\theta)).
This expresses the target quantity in terms of known sum and product values.
4
Substitute the values of the sum and product into the factored expression and simplify.
sin⁡3(θ)+cos⁡3(θ)=(72)(1−38)=(72)(58)=5716\sin^3(\theta) + \cos^3(\theta) = \left(\frac{\sqrt{7}}{2}\right)\left(1 - \frac{3}{8}\right) = \left(\frac{\sqrt{7}}{2}\right)\left(\frac{5}{8}\right) = \frac{5\sqrt{7}}{16}.
This yields the final simplified numerical ratio.

Key Concept

Pythagorean Identity and Algebraic Manipulation of Trigonometric Ratios
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Topics

Trigonometric Ratios and Identities

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