Discussion
Considering only the principal values of the inverse trigonometric functions, the value of
\(\cot^{-1}(\cot(-11)) + 10\sin(2\cos^{-1}(\frac{1}{\sqrt{2}}))\)
\(+\, 10\sin(2\tan^{-1}(2))\)
is
Answer: C
The value is \(4\pi + 7\).
The principal value of \(\cot^{-1}\) lies in \((0, \pi)\), and cot has period \(\pi\), so add multiples of \(\pi\) to \(-11\) until it lands in that interval: \(-11 + 4\pi \approx 1.566\), which lies in \((0, \pi)\). So \(\cot^{-1}(\cot(-11)) = 4\pi - 11\).
\[ \begin{aligned} 10\sin\left(2 \cdot \tfrac{\pi}{4}\right) &= 10 \\ \sin(2\tan^{-1} 2) &= \frac{2 \cdot 2}{1 + 2^{2}} = \frac{4}{5} \\ \text{total} &= (4\pi - 11) + 10 + 8 = 4\pi + 7 \end{aligned} \]
The answer 7 comes from writing \(\cot^{-1}(\cot(-11)) = -11\), forgetting the principal range. \(3\pi + 7\) uses \(3\pi - 11\), which is negative and so also outside \((0, \pi)\).
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