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Q14:
$\sin^{-1}(1 - x) - 2\sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to
Solution :
Step 1: Establishing the Domain and Substitution
Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1} x = \frac{\pi}{2}$
We restrict the variable $x$ based on the domain of the inverse sine function, $\sin^{-1} \theta$, where $\theta \in [-1, 1]$. Thus:
- $1 - x \in [-1, 1] \implies x \in [0, 2]$
- $x \in [-1, 1]$
Step 2: Rearranging the Equation
Isolate the term $\sin^{-1}(1-x)$: $$\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1} x$$ Apply the sine function to both sides: $$1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right)$$ [Using the trigonometric identity $\sin(\frac{\pi}{2} + \theta) = \cos \theta$]: $$1 - x = \cos(2\sin^{-1} x)$$
Step 3: Algebraic Substitution and Expansion
Let $\theta = \sin^{-1} x$, which implies $\sin \theta = x$. Substituting this into the identity $\cos(2\theta) = 1 - 2\sin^2 \theta$: $$1 - x = 1 - 2\sin^2(\sin^{-1} x)$$ $$1 - x = 1 - 2x^2$$ [By the property $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$].
Step 4: Solving the Quadratic Equation
Subtract 1 from both sides: $$-x = -2x^2$$ Rearrange to standard quadratic form: $$2x^2 - x = 0$$ Factor the expression: $$x(2x - 1) = 0$$ This yields two potential roots:
- $x_1 = 0$
- $x_2 = \frac{1}{2}$
Step 5: Verification of Roots
We test both roots against the original equation to ensure they do not produce extraneous solutions:
| Value | LHS Calculation | Result |
|---|---|---|
| $x = 0$ | $\sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2}$ | Valid ($\frac{\pi}{2} = \frac{\pi}{2}$) |
| $x = \frac{1}{2}$ | $\sin^{-1}(1 - \frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2(\frac{\pi}{6}) = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6}$ | Invalid ($-\frac{\pi}{6} \neq \frac{\pi}{2}$) |
Visual Representation of the Functions
Final Solution: x = 0
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.MISC
- Q1: Find the value of the following: $\cos^{-1}(\cos(\frac{13\pi}{6}))$
- Q2: Find the value of the following: $\tan^{-1}(\tan(\frac{7\pi}{6}))$
- Q3: Prove that $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$
- Q4: Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$
- Q5: Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$
- Q6: Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
- Q7: Prove that $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
- Q8: Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x})$, $x \in [0, 1]$
- Q9: Prove that $\cot^{-1} (\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}) = \frac{x}{2}$, $x \in (0, \frac{\pi}{4})$
- Q10: Prove that $\tan^{-1}(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$, $\frac{-1}{\sqrt{2}} \leq x \leq 1$
- Q11: Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$
- Q12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x > 0)$
- Q13: $\sin(\tan^{-1} x)$, $|x| < 1$ is equal to
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