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Difficulty: Very hardTrigonometric Ratios and Identities

For an acute angle θ\theta, the expression sin⁡4(θ)−cos⁡4(θ)sin⁡(θ)−cos⁡(θ)\frac{\sin^4(\theta) - \cos^4(\theta)}{\sin(\theta) - \cos(\theta)} is equal to 75\frac{7}{5}. What is the value of 25sin⁡(θ)cos⁡(θ)25\sin(\theta)\cos(\theta)?

Answer: 12

Answer

The correct answer is 12.
The expression sin⁡4(θ)−cos⁡4(θ)sin⁡(θ)−cos⁡(θ)\frac{\sin^4(\theta) - \cos^4(\theta)}{\sin(\theta) - \cos(\theta)} simplifies directly to sin⁡(θ)+cos⁡(θ)=75\sin(\theta) + \cos(\theta) = \frac{7}{5} by applying the difference of squares identity twice and substituting the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1. Squaring both sides of this simplified equation gives sin⁡2(θ)+2sin⁡(θ)cos⁡(θ)+cos⁡2(θ)=4925\sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = \frac{49}{25}. Replacing sin⁡2(θ)+cos⁡2(θ)\sin^2(\theta) + \cos^2(\theta) with 1 yields 1+2sin⁡(θ)cos⁡(θ)=49251 + 2\sin(\theta)\cos(\theta) = \frac{49}{25}, which simplifies to sin⁡(θ)cos⁡(θ)=1225\sin(\theta)\cos(\theta) = \frac{12}{25}. Multiplying this result by 25 gives the final integer value of 12.

Step-by-Step Solution

1
Factor the numerator of the expression.
sin⁡4(θ)−cos⁡4(θ)=(sin⁡2(θ)−cos⁡2(θ))(sin⁡2(θ)+cos⁡2(θ))\sin^4(\theta) - \cos^4(\theta) = (\sin^2(\theta) - \cos^2(\theta))(\sin^2(\theta) + \cos^2(\theta))
The difference of squares identity can be applied to terms with fourth powers.
2
Apply the Pythagorean identity to simplify the factored numerator.
sin⁡4(θ)−cos⁡4(θ)=sin⁡2(θ)−cos⁡2(θ)\sin^4(\theta) - \cos^4(\theta) = \sin^2(\theta) - \cos^2(\theta)
The Pythagorean trigonometric identity states that sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1.
3
Factor the remaining term in the numerator.
sin⁡2(θ)−cos⁡2(θ)=(sin⁡(θ)−cos⁡(θ))(sin⁡(θ)+cos⁡(θ))\sin^2(\theta) - \cos^2(\theta) = (\sin(\theta) - \cos(\theta))(\sin(\theta) + \cos(\theta))
This is another application of the difference of squares identity.
4
Simplify the fraction by dividing the common factor in the numerator and denominator.
sin⁡(θ)+cos⁡(θ)=75\sin(\theta) + \cos(\theta) = \frac{7}{5}
The term sin⁡(θ)−cos⁡(θ)\sin(\theta) - \cos(\theta) in the numerator and denominator cancels out since θ\theta is an acute angle and sin⁡(θ)≠cos⁡(θ)\sin(\theta) \neq \cos(\theta).
5
Square both sides of the simplified equation.
sin⁡2(θ)+2sin⁡(θ)cos⁡(θ)+cos⁡2(θ)=4925\sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = \frac{49}{25}
Squaring both sides allows us to relate the sum sin⁡(θ)+cos⁡(θ)\sin(\theta) + \cos(\theta) to the product sin⁡(θ)cos⁡(θ)\sin(\theta)\cos(\theta).
6
Substitute the Pythagorean identity and solve for the product of sine and cosine.
sin⁡(θ)cos⁡(θ)=1225\sin(\theta)\cos(\theta) = \frac{12}{25}
Substituting sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 yields 1+2sin⁡(θ)cos⁡(θ)=49251 + 2\sin(\theta)\cos(\theta) = \frac{49}{25}, which simplifies to 2sin⁡(θ)cos⁡(θ)=24252\sin(\theta)\cos(\theta) = \frac{24}{25}.
7
Multiply the product by 25 to find the target value.
12
The question asks for the value of 25sin⁡(θ)cos⁡(θ)25\sin(\theta)\cos(\theta).

Key Concept

Simplifying trigonometric expressions using algebraic factorization and Pythagorean identities.

Alternative Method

Since θ\theta is an acute angle in a right triangle, we can test standard Pythagorean triples. A right triangle with side lengths 3, 4, and 5 has an angle θ\theta where sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and cos⁡(θ)=45\cos(\theta) = \frac{4}{5}. Checking these values in the simplified expression gives sin⁡(θ)+cos⁡(θ)=35+45=75\sin(\theta) + \cos(\theta) = \frac{3}{5} + \frac{4}{5} = \frac{7}{5}, which matches the given condition. We can then directly calculate 25sin⁡(θ)cos⁡(θ)=25(35)(45)=1225\sin(\theta)\cos(\theta) = 25 \left(\frac{3}{5}\right)\left(\frac{4}{5}\right) = 12.
Estimated Time:3m 0s
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Topics

Trigonometric Ratios and Identities

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