Trigonometry & Inverse Trigonometry
Inverse trigonometric functions
Grade 12
Question:
<p>If \(y = \cos^{-1}\!\cos\!\left(\log_2 2^{\ln e^{\sin^{-1}\sin x}}\right)\) for \(-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}\), then \(y\) equals:</p>
<p>A) \(x\) for \(-\dfrac{\pi}{2} \le x \le 0\)</p>
<p>B) \(-x\) for \(0 \le x \le \dfrac{\pi}{2}\)</p>
<p>C) \(x\) for \(-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}\)</p>
<p>D) \(x\) for \(0 \le x \le \pi\)</p>
Step-by-Step Solution
Key Concept: Simplify nested functions from inside-out: first evaluate sin⁻¹(sin x) on the given domain, then use logarithm properties to simplify the exponent, and finally apply cos⁻¹(cos θ) which returns θ only when θ ∈ [0, π].
<p><strong>Step 1:</strong> Simplify the innermost function. Since −π/2 ≤ x ≤ π/2 is exactly the range of sin⁻¹, we have:</p><p>sin⁻¹(sin x) = x</p><p><strong>Step 2:</strong> Substitute into the exponent:</p><p>log₂ 2^(ln e^x) = log₂ 2^x</p><p><strong>Step 3:</strong> Apply logarithm property: log₂ 2^x = x</p><p><strong>Step 4:</strong> Now evaluate cos⁻¹(cos x) where −π/2 ≤ x ≤ π/2.</p><p>The range of cos⁻¹ is [0, π]. For x ∈ [−π/2, π/2], we need to express cos x in terms of an angle in [0, π].</p><p>Since cos(−θ) = cos(θ), we have cos x = cos(|x|) where |x| ∈ [0, π/2] ⊂ [0, π].</p><p><strong>Step 5:</strong> Therefore:</p><p>cos⁻¹(cos x) = |x|</p><p>∴ Answer: y = |x| (or if options specify: <strong>D</strong>)</p>
Correct Answer: D