Q111 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { Evaluate: } \cos ^{-1}\left(\cos \dfrac{35 \pi}{18}\right)-\sin ^{-1}\left(\sin \dfrac{35 \pi}{18}\right)$
- Correct answerA$\dfrac{\pi}{9}$
- B$\pi$
- C0
- D$\dfrac{\pi}{18}$
Solution
The expression is $\cos ^{-1}\left(\cos \dfrac{35 \pi}{18}\right)-\sin ^{-1}\left(\sin \dfrac{35 \pi}{18}\right)$.
First, simplify the argument $\dfrac{35 \pi}{18}$ as $2\pi - \dfrac{\pi}{18}$.
For the cosine term: $\cos ^{-1}\left(\cos \left(2\pi - \dfrac{\pi}{18}\right)\right) = \cos ^{-1}\left(\cos \dfrac{\pi}{18}\right) = \dfrac{\pi}{18}$, since $\dfrac{\pi}{18} \in [0, \pi]$.
For the sine term: $\sin ^{-1}\left(\sin \left(2\pi - \dfrac{\pi}{18}\right)\right) = \sin ^{-1}\left(\sin \left(-\dfrac{\pi}{18}\right)\right) = -\dfrac{\pi}{18}$, since $-\dfrac{\pi}{18} \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$.
Substituting these values back into the expression:
$\dfrac{\pi}{18} - \left(-\dfrac{\pi}{18}\right) = \dfrac{\pi}{18} + \dfrac{\pi}{18} = \dfrac{2\pi}{18} = \dfrac{\pi}{9}$.
Answer: $\dfrac{\pi}{9}$
First, simplify the argument $\dfrac{35 \pi}{18}$ as $2\pi - \dfrac{\pi}{18}$.
For the cosine term: $\cos ^{-1}\left(\cos \left(2\pi - \dfrac{\pi}{18}\right)\right) = \cos ^{-1}\left(\cos \dfrac{\pi}{18}\right) = \dfrac{\pi}{18}$, since $\dfrac{\pi}{18} \in [0, \pi]$.
For the sine term: $\sin ^{-1}\left(\sin \left(2\pi - \dfrac{\pi}{18}\right)\right) = \sin ^{-1}\left(\sin \left(-\dfrac{\pi}{18}\right)\right) = -\dfrac{\pi}{18}$, since $-\dfrac{\pi}{18} \in [-\dfrac{\pi}{2}, \dfrac{\pi}{2}]$.
Substituting these values back into the expression:
$\dfrac{\pi}{18} - \left(-\dfrac{\pi}{18}\right) = \dfrac{\pi}{18} + \dfrac{\pi}{18} = \dfrac{2\pi}{18} = \dfrac{\pi}{9}$.
Answer: $\dfrac{\pi}{9}$
Hence, the correct answer is option: 'A'.