Discussion
if 5 \(sin^{-1}\alpha + 3cos^{-1}\alpha = \pi\) ,then \(\alpha\) is equal to
Answer: C
\(\alpha = \) \(\frac{-1}{\sqrt{2}}\).
Use \(\cos^{-1}\alpha = \frac{\pi}{2} - \sin^{-1}\alpha\):
\[ \begin{aligned} 5\sin^{-1}\alpha + \tfrac{3\pi}{2} - 3\sin^{-1}\alpha &= \pi \\ 2\sin^{-1}\alpha &= -\tfrac{\pi}{2} \\ \sin^{-1}\alpha &= -\tfrac{\pi}{4} \end{aligned} \]
So \(\alpha = \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}\). Check: \(5(-\frac{\pi}{4}) + 3(\frac{3\pi}{4}) = \pi\). \(\frac{1}{\sqrt{2}}\) loses the sign; 1 and -1 give \(\frac{5\pi}{2}\) and \(\frac{\pi}{2}\) instead.
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