Mathematics

Mathematics

Problem

(a) Prove that arcsin⁡(−x)=−arcsin⁡x\arcsin(-x)=-\arcsin x for every x∈[−1,1]x\in[-1,1]. (b) Using the result of part (a), evaluate the following, giving each answer as a principal value: (i) arcsin⁡(sin⁡(−11π6))\arcsin\bigl(\sin\bigl(-\dfrac{11\pi}{6}\bigr)\bigr) (ii) arcsin⁡(sin⁡(−5π4))\arcsin\bigl(\sin\bigl(-\dfrac{5\pi}{4}\bigr)\bigr) (c) Explain briefly why the answers obtained in part (b) lie in the principal value range of arcsin⁡x\arcsin x.
Intuitive explanation

The key insight is that inverse trigonometric functions have symmetry properties rooted in the symmetry of their parent functions. Since sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta), the arcsine function inherits an odd symmetry: if you negate the input, you negate the output. Part (a) formalizes this algebraically using the definition of inverse functions. Once you've established this property, parts (b) and (c) become about carefully reducing angles outside the principal range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] down to equivalent sine values inside it, then applying the symmetry to simplify.

Intuitive explanation

The key insight is that inverse trigonometric functions have symmetry properties rooted in the symmetry of their parent functions. Since sin⁡(−θ)=−sin⁡(θ)\sin(-\theta) = -\sin(\theta), the arcsine function inherits an odd symmetry: if you negate the input, you negate the output. Part (a) formalizes this algebraically using the definition of inverse functions. Once you've established this property, parts (b) and (c) become about carefully reducing angles outside the principal range [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] down to equivalent sine values inside it, then applying the symmetry to simplify.

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