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Difficulty: HardTrigonometric Ratios and Identities

For an acute angle θ\theta, the equation 2cos⁡2(θ)−5sin⁡(θ)+1=02\cos^2(\theta) - 5\sin(\theta) + 1 = 0 is true. What is the value of sin⁡(θ)\sin(\theta)?

Answer: 0.5

Answer

The value of sin⁡(θ)\sin(\theta) is 0.50.5 (or 12\frac{1}{2})
By applying the Pythagorean identity cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta), the equation 2cos⁡2(θ)−5sin⁡(θ)+1=02\cos^2(\theta) - 5\sin(\theta) + 1 = 0 can be rewritten entirely in terms of sin⁡(θ)\sin(\theta), yielding −2sin⁡2(θ)−5sin⁡(θ)+3=0-2\sin^2(\theta) - 5\sin(\theta) + 3 = 0. Multiplying by −1-1 gives the standard quadratic equation 2sin⁡2(θ)+5sin⁡(θ)−3=02\sin^2(\theta) + 5\sin(\theta) - 3 = 0, which factors as (2sin⁡(θ)−1)(sin⁡(θ)+3)=0(2\sin(\theta) - 1)(\sin(\theta) + 3) = 0. Solving for sin⁡(θ)\sin(\theta) yields sin⁡(θ)=0.5\sin(\theta) = 0.5 or sin⁡(θ)=−3\sin(\theta) = -3. Since the sine value of any angle must be in the range [−1,1][-1, 1] and the sine of an acute angle must be positive, sin⁡(θ)=0.5\sin(\theta) = 0.5 is the only valid solution.

Step-by-Step Solution

1
Apply the Pythagorean identity to rewrite the cosine term.
2(1−sin⁡2(θ))−5sin⁡(θ)+1=02(1 - \sin^2(\theta)) - 5\sin(\theta) + 1 = 0
The equation contains both cos⁡2(θ)\cos^2(\theta) and sin⁡(θ)\sin(\theta). Substituting cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta) allows the equation to be expressed in terms of a single trigonometric function, sin⁡(θ)\sin(\theta).
2
Distribute and simplify the equation into standard quadratic form.
2sin⁡2(θ)+5sin⁡(θ)−3=02\sin^2(\theta) + 5\sin(\theta) - 3 = 0
Expanding the equation yields 2−2sin⁡2(θ)−5sin⁡(θ)+1=02 - 2\sin^2(\theta) - 5\sin(\theta) + 1 = 0, which simplifies to −2sin⁡2(θ)−5sin⁡(θ)+3=0-2\sin^2(\theta) - 5\sin(\theta) + 3 = 0. Multiplying the entire equation by −1-1 puts it into standard quadratic form as2+bs+c=0as^2 + bs + c = 0.
3
Factor the quadratic expression.
(2sin⁡(θ)−1)(sin⁡(θ)+3)=0(2\sin(\theta) - 1)(\sin(\theta) + 3) = 0
Finding two numbers that multiply to −6-6 (from 2×−32 \times -3) and add to 55 leads to the factors 66 and −1-1. Splitting the middle term and factoring by grouping yields (2sin⁡(θ)−1)(sin⁡(θ)+3)=0(2\sin(\theta) - 1)(\sin(\theta) + 3) = 0.
4
Determine the valid solution based on the angle's constraints.
sin⁡(θ)=0.5\sin(\theta) = 0.5
Setting each factor to zero gives sin⁡(θ)=0.5\sin(\theta) = 0.5 or sin⁡(θ)=−3\sin(\theta) = -3. Since the sine of any real angle must be between −1-1 and 11, sin⁡(θ)=−3\sin(\theta) = -3 is undefined. Furthermore, because θ\theta is an acute angle (0∘<θ<90∘0^\circ < \theta < 90^\circ), the sine value must be positive, confirming sin⁡(θ)=0.5\sin(\theta) = 0.5.

Key Concept

Pythagorean identity and quadratic trigonometric equations
Estimated Time:2m 0s
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Topics

Trigonometric Ratios and Identities

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