The value of $\cos(2\cos^{-1}x + \sin^{-1}x)$ at $x = \frac{1}{5}$ is

The expression to evaluate is $\cos(2\cos^{-1}x + \sin^{-1}x)$ at $x = \frac{1}{5}$.

Using the identity $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$:

$2\cos^{-1}x + \sin^{-1}x = \cos^{-1}x + \cos^{-1}x + \sin^{-1}x$

$= \cos^{-1}x + (\cos^{-1}x + \sin^{-1}x)$

$= \cos^{-1}x + \frac{\pi}{2}$

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Applying the cosine shift formula $\cos\left(A + \frac{\pi}{2}\right) = -\sin(A)$:

$\cos\left(\cos^{-1}x + \frac{\pi}{2}\right) = -\sin(\cos^{-1}x)$

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Let $\cos^{-1}x = \theta$, then $\cos\theta = x$.

Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:

$\sin\theta = \sqrt{1 - \cos^2\theta}$

$= \sqrt{1 - x^2}$

Therefore $\sin(\cos^{-1}x) = \sqrt{1-x^2}$.

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The expression simplifies to:

$\cos(2\cos^{-1}x + \sin^{-1}x) = -\sqrt{1-x^2}$

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Substituting $x = \frac{1}{5}$:

$= -\sqrt{1 - \left(\frac{1}{5}\right)^2}$

$= -\sqrt{1 - \frac{1}{25}}$

$= -\sqrt{\frac{24}{25}}$

Therefore, the value is $-\sqrt{\frac{24}{25}}$.