Question
Show that:
\[ \sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x \]
Solution
Let
\[ \sin^{-1}x = \theta \]
Then,
\[ x = \sin \theta \]
So,
\[ \sqrt{1 – x^2} = \sqrt{1 – \sin^2\theta} = \cos \theta \]
Now substitute in LHS:
\[ \sin^{-1}\left(2x\sqrt{1-x^2}\right) = \sin^{-1}(2\sin\theta \cos\theta) \]
Using identity:
\[ 2\sin\theta \cos\theta = \sin 2\theta \]
Thus,
\[ = \sin^{-1}(\sin 2\theta) \]
Now, if \( \theta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \), then \( 2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)
So,
\[ \sin^{-1}(\sin 2\theta) = 2\theta \]
Hence,
\[ = 2\sin^{-1}x \]
Final Result:
\[ \boxed{ \sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x } \]
Key Concept
Use substitution \( x = \sin\theta \) and identity \( \sin 2\theta = 2\sin\theta\cos\theta \), along with principal value conditions.