Question
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Write the expression in algebraic form. (Hint: Sketch a right triangle.) \[ \cos (\arcsin (5 x)) \]

Ask by Huang Lowe. in the United States
Feb 26,2025

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Tutor-Verified Answer

Answer

cos(arcsin(5x)) = √(1 - 25x²), where |x| ≤ 1/5.

Solution

Let θ = arcsin(5x). Then sin θ = 5x. By the Pythagorean identity, we have:   cos²θ + sin²θ = 1   cos²θ = 1 - sin²θ = 1 - (5x)² = 1 - 25x² Since the range of arcsin is [–π/2, π/2], where cosine is nonnegative, we have:   cos θ = √(1 - 25x²) Thus, cos(arcsin(5x)) = √(1 - 25x²), provided that |5x| ≤ 1 (so |x| ≤ 1/5).

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Bonus Knowledge

To express \( \cos (\arcsin (5 x)) \) in algebraic form, consider a right triangle where the angle \( \theta = \arcsin(5x) \). Here, the opposite side is \( 5x \) and the hypotenuse is \( 1 \) (since \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)). By using the Pythagorean theorem, the adjacent side can be calculated as \( \sqrt{1^2 - (5x)^2} = \sqrt{1 - 25x^2} \). Thus, the cosine of the angle \( \theta \) can be expressed as: \[ \cos (\arcsin (5x)) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{1 - 25x^2}}{1} = \sqrt{1 - 25x^2}. \] Next, remember that \( 5x \) must stay within the range [-1, 1] for the arcsine function to be defined, so ensure \( -\frac{1}{5} \leq x \leq \frac{1}{5} \)!

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