$\cos\!\left(\sin^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{5}{13}+\sin^{-1}\dfrac{33}{65}\right)$ is equal to:
Step-by-Step Solution
Key Concept: Convert each $\sin^{-1}$ to $\tan^{-1}$, combine the first two using the addition formula, then recognise the sum of the result and the third term equals $\tfrac{\pi}{2}$.
$\sin^{-1}\tfrac{3}{5}=\tan^{-1}\tfrac{3}{4}$,\quad $\sin^{-1}\tfrac{5}{13}=\tan^{-1}\tfrac{5}{12}$,\quad $\sin^{-1}\tfrac{33}{65}=\tan^{-1}\tfrac{33}{56}$.
$\tan^{-1}\tfrac{3}{4}+\tan^{-1}\tfrac{5}{12}=\tan^{-1}\!\dfrac{\tfrac{3}{4}+\tfrac{5}{12}}{1-\tfrac{3}{4}\cdot\tfrac{5}{12}}=\tan^{-1}\!\dfrac{\tfrac{56}{48}}{\tfrac{33}{48}}=\tan^{-1}\dfrac{56}{33}$.
Then $\tan^{-1}\dfrac{56}{33}+\tan^{-1}\dfrac{33}{56}=\tan^{-1}\dfrac{56}{33}+\cot^{-1}\dfrac{56}{33}=\dfrac{\pi}{2}$.
Therefore the argument of $\cos$ is $\dfrac{\pi}{2}$, so the answer is $\cos\dfrac{\pi}{2}=0$.
Correct Answer: 2