Problem 20 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve for \(x: \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)+\cos ^{-1} x=\frac{\pi}{4}\).

Step-by-Step Solution

Answer

The solution to the given equation is \( x = \sqrt{0.7} \)

Step 1: Write down the equation

Start by writing the equation: \(\sin^{-1} \left( \frac{1}{\sqrt{5}} \right) + \cos^{-1} x = \frac{\pi}{4}\)

Step 2: Isolate the inverse cosine term

Subtract \(\sin^{-1} \left( \frac{1}{\sqrt{5}} \right)\) from both sides of the equation to isolate the term involving \( x \): \(\cos^{-1} x = \frac{\pi}{4} - \sin^{-1} \left( \frac{1}{\sqrt{5}} \right)\)

Step 3: Apply the trigonometric identity

Apply the trigonometric identity \(\sin^{2}(θ) + \cos^{2}(θ) = 1\) and remember that \(\cos^{2}(θ) = 1 - \sin^{2}(θ)\), so that \(\cos θ = ±\sqrt{1 - \sin^{2}(θ)}\). From this, we have \(x = ±\sqrt{1 - \sin^{2} \left(\frac{\pi}{4} - \sin^{-1} \left( \frac{1}{\sqrt{5}} \right)\right)}\)

Step 4: Simplify the equation

Significant simplification becomes possible if we remember that the square of the inverse sine term \(\sin^{-1} \left( \frac{1}{\sqrt{5}} \right)\) is \(\left( \frac{1}{\sqrt{5}} \right)^{2}\), which equals \( \frac{1}{5} \). Therefore, \(x = ±\sqrt{1 - \left(\frac{1}{2} - \frac{1}{5}\right)} = ±\sqrt{\frac{7}{10}} = ±\sqrt{0.7}\).

Step 5: Determine the range and domain

Recall that the domain (possible values of \( x \)) for the inverse cosine term is such that \( -1 \leq x \leq 1 \). \(\sqrt{0.7}\) is a valid answer falling within the domain, however, \(-\sqrt{0.7}\) makes \(\cos^{-1}x\) undefined. Hence, only one solution exists: \(x = \sqrt{0.7}\).