Find the value of : cos(2 cos⁻¹ x +...

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1981Easy

Find the value of : $cos (2 cos^{- 1} x + sin^{- 1} x)$ at $x = \frac{1}{5}$ , where $0 \leq cos^{- 1} x \leq π$ and $- π /2 \leq sin^{- 1} x \leq π /2$ .

Answer:

- \frac{2 6}{5}

Visualized Solution (Hindi)

Analyze the Expression

Given expression: $cos (2 cos^{- 1} x + sin^{- 1} x)$
At $x = \frac{1}{5}$
Step 1: Split the term $2 cos^{- 1} x$
$cos (cos^{- 1} x + cos^{- 1} x + sin^{- 1} x)$

Apply Inverse Identity

Recall the identity: $cos^{- 1} x + sin^{- 1} x = \frac{π}{2}$
Substitute this into the expression:
$cos (cos^{- 1} x + \frac{π}{2})$

Trigonometric Reduction

Use the allied angle formula: $cos (\frac{π}{2} + θ) = - sin θ$
Here, $θ = cos^{- 1} x$
Expression becomes: $- sin (cos^{- 1} x)$

Visualizing the Triangle

Let $θ = cos^{- 1} x ⟹ cos θ = x$
In a right triangle: $Base = x$ , $Hypotenuse = 1$
$Perpendicular = 1 - x^{2}$
Therefore, $sin (cos^{- 1} x) = 1 - x^{2}$
Expression: $- 1 - x^{2}$

Final Substitution and Result

Substitute $x = \frac{1}{5}$ :
$- 1 - (\frac{1}{5})^{2} = - 1 - \frac{1}{25}$
$- \frac{24}{25} = - \frac{24}{5}$
Final Answer: $- \frac{2 6}{5}$

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the value of : cos(2cos−1x+sin−1x) at x=51​, where 0≤cos−1x≤π and −π/2≤sin−1x≤π/2.

Visualized Solution (Hindi)

Analyze the Expression

Apply Inverse Identity

Trigonometric Reduction

Visualizing the Triangle

Final Substitution and Result

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