If y = 3 sin^{-1}x + sin^{-1}(3x - 4x^3) for | vedclass

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(A) Let $x = \sin	heta$. Since $x \in [-1/2, 1/2]$,we have $	heta \in [-\pi/6, \pi/6]$.The expression becomes $y = 3\sin^{-1}(\sin	heta) + \sin^{-1

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$-\pi \leq y \leq \pi$

Vedclass · Answer

(A) Let $x = \sin	heta$. Since $x \in [-1/2, 1/2]$,we have $	heta \in [-\pi/6, \pi/6]$.The expression becomes $y = 3\sin^{-1}(\sin	heta) + \sin^{-1}(\sin(3	heta))$.Since $	heta \in [-\pi/6, \pi/6]$,we have $3	heta \in [-\pi/2, \pi/2]$.Therefore,$\sin^{-1}(\sin	heta) = 	heta$ and $\sin^{-1}(\sin(3	heta)) = 3	heta$.Substituting these,we get $y = 3	heta + 3	heta = 6	heta$.Given $	heta \in [-\pi/6, \pi/6]$,multiplying by $6$ gives $6	heta \in [-\pi, \pi]$.Thus,$y \in [-\pi, \pi]$.

If $y = 3 \sin^{-1}x + \sin^{-1}(3x - 4x^3)$ for all $x \in [-1/2, 1/2]$,then

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