We are given an equation involving inverse trigonometric functions: \(4 \sin^{-1} x + \cos^{-1} x = \pi\) Our goal is to find the value of the expression \(\sin^{-1} x + 4 \cos^{-1} x\).
We will use the fundamental identity relating the inverse sine and inverse cosine functions: \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) This identity holds true for all \(x\) in the domain \([-1, 1]\).
Let's manipulate the given equation using the identity. We can rewrite \(4 \sin^{-1} x\) as \(3 \sin^{-1} x + \sin^{-1} x\).
The given equation becomes: \( (3 \sin^{-1} x) + (\sin^{-1} x + \cos^{-1} x) = \pi \)
Now, substitute the identity \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\): \( 3 \sin^{-1} x + \frac{\pi}{2} = \pi \)
Solve for \(\sin^{-1} x\): \( 3 \sin^{-1} x = \pi - \frac{\pi}{2} \) \( 3 \sin^{-1} x = \frac{\pi}{2} \) \( \sin^{-1} x = \frac{\pi}{6} \)
Now, find \(\cos^{-1} x\) using the identity: \( \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \) \( \cos^{-1} x = \frac{\pi}{2} - \frac{\pi}{6} \) \( \cos^{-1} x = \frac{3\pi}{6} - \frac{\pi}{6} \) \( \cos^{-1} x = \frac{2\pi}{6} = \frac{\pi}{3} \)
We need to find the value of \(\sin^{-1} x + 4 \cos^{-1} x\). Substitute the values we found for \(\sin^{-1} x\) and \(\cos^{-1} x\): \( \sin^{-1} x + 4 \cos^{-1} x = \left( \frac{\pi}{6} \right) + 4 \left( \frac{\pi}{3} \right) \)
Simplify the expression: \( = \frac{\pi}{6} + \frac{4\pi}{3} \) \( = \frac{\pi}{6} + \frac{8\pi}{6} \) \( = \frac{\pi + 8\pi}{6} \) \( = \frac{9\pi}{6} \) \( = \frac{3\pi}{2} \)
Thus, the value of the expression \(\sin^{-1} x + 4 \cos^{-1} x\) is \( \frac{3\pi}{2} \).
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