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Exercise 4.6 · Q5

Q.sin⁡−1(cos⁡x)=π2−x\sin^{-1}(\cos x) = \dfrac{\pi}2 - x is valid for

(1) −π≤x≤0-\pi \le x \le 0
(2) 0≤x≤π0 \le x \le \pi
(3) −π2≤x≤π2-\dfrac{\pi}2 \le x \le \dfrac{\pi}2
(4) −π4≤x≤3π4-\dfrac{\pi}4 \le x \le \dfrac{3\pi}4
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The identity sin⁡−1(cos⁡x)=π2−x\sin^{-1}(\cos x)=\dfrac{\pi}2-x requires the right side to land in [−π2,π2]\left[-\dfrac{\pi}2,\dfrac{\pi}2\right], which pins down exactly which xx make it valid.

Step 1. State the requirement. For sin⁡−1(cos⁡x)=π2−x\sin^{-1}(\cos x)=\dfrac{\pi}2-x to hold, we need π2−x∈[−π2,π2]\dfrac{\pi}2-x\in\left[-\dfrac{\pi}2,\dfrac{\pi}2\right]. …

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