Question

The value of $\cos(\cos^{-1}\frac{1}{5}+2\sin^{-1}\frac{1}{5})$ is equal to

• A. $\frac{4}{5}$
• B. $\frac{-4}{5}$
• C. $\frac{3}{5}$
• D. $\frac{-1}{5}$
• E. $\frac{1}{5}$

Hint:

Formula Tip: Always look for complementary pairs! If you see a mix of $\sin^{-1}(x)$ and $\cos^{-1}(x)$ with the same argument '$x$', group them immediately to form a $90^\circ$ angle.

Answer 1

The Correct Option is D

Concept:
Inverse trigonometric functions possess complementary identities. Specifically, for any value $x \in [-1, 1]$, the sum of the inverse sine and inverse cosine is always a right angle: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$. Furthermore, the standard identity $\cos(\frac{\pi}{2} + \theta) = -\sin\theta$ is used for simplification.

Step 1: State the given expression.
We are given the trigonometric expression: $$E = \cos\left(\cos^{-1}\frac{1}{5} + 2\sin^{-1}\frac{1}{5}\right)$$

Step 2: Split the inner term to isolate the complementary pair.
We have $2\sin^{-1}\frac{1}{5}$. We can break this into two separate terms to pair one up with the inverse cosine: $$E = \cos\left( \left[\cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5}\right] + \sin^{-1}\frac{1}{5} \right)$$

Step 3: Apply the complementary identity.
Using the property $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, the bracketed portion evaluates to $\frac{\pi}{2}$: $$E = \cos\left( \frac{\pi}{2} + \sin^{-1}\frac{1}{5} \right)$$

Step 4: Apply the quadrant angle transformation.
Let $\theta = \sin^{-1}\frac{1}{5}$. The expression is now in the form $\cos(\frac{\pi}{2} + \theta)$. Since adding $90^\circ$ ($\frac{\pi}{2}$) moves the angle into the second quadrant where cosine is negative, the identity is $\cos(\frac{\pi}{2} + \theta) = -\sin\theta$. $$E = -\sin\left( \sin^{-1}\frac{1}{5} \right)$$

Step 5: Evaluate the inverse function composition.
The sine function and the inverse sine function cancel each other out, $\sin(\sin^{-1}x) = x$: $$E = -\left(\frac{1}{5}\right)$$ $$E = \frac{-1}{5}$$ Hence the correct answer is (D) $\frac{-1{5}$}.