Question
Let
\[ f(x) = e^{\cos^{-1}(\sin(x + \tfrac{\pi}{3}))} \]
Find \( f\left(\tfrac{8\pi}{9}\right) \).
Solution
Substitute:
\[ x + \frac{\pi}{3} = \frac{8\pi}{9} + \frac{\pi}{3} = \frac{8\pi}{9} + \frac{3\pi}{9} = \frac{11\pi}{9} \]
Now,
\[ \sin\left(\frac{11\pi}{9}\right) = \sin\left(\pi + \frac{2\pi}{9}\right) = -\sin\left(\frac{2\pi}{9}\right) \]
So expression becomes:
\[ \cos^{-1}\left(-\sin\left(\frac{2\pi}{9}\right)\right) \]
Use identity:
\[ \sin\theta = \cos\left(\frac{\pi}{2} – \theta\right) \]
\[ -\sin\left(\frac{2\pi}{9}\right) = -\cos\left(\frac{\pi}{2} – \frac{2\pi}{9}\right) = -\cos\left(\frac{5\pi}{18}\right) \]
\[ = \cos\left(\pi – \frac{5\pi}{18}\right) = \cos\left(\frac{13\pi}{18}\right) \]
Thus,
\[ \cos^{-1}\left(\cos\frac{13\pi}{18}\right) = \frac{13\pi}{18} \]
(since it lies in principal range [0, π])
Therefore,
\[ f\left(\frac{8\pi}{9}\right) = e^{\frac{13\pi}{18}} \]
Final Answer:
\[ \boxed{e^{\frac{13\pi}{18}}} \]
Key Concept
Convert sin into cos form and apply principal value of cos⁻¹ carefully.