Solving an Equation Involving Inverse Trigonometric Functions Solve [latex]\cos^{-1} x = \sin^{-1} \frac{1}{2}[/latex] .

Let [latex]\sin^{-1} \frac{1}{2} = u[/latex]. Then [latex]\sin u = \frac{1}{2}[/latex] , and for u in quadrant I we have the following. [latex]\cos^{-1} x = \sin^{-1} \frac{1}{2}[/latex] Original equation [latex]\cos^{-1} x = u[/latex] Substitute. [latex]\cos u = x[/latex] Alternative form Sketch a triangle and label it using the facts that u is in quadrant I and [latex]\sin u = \frac{1}{2}[/latex]. See Figure 39. Because [latex]x = \cos u[/latex], we have [latex]x = \frac{\sqrt{3}}{2}[/latex] . The solution set is {[latex] \frac{\sqrt{3}}{2}[/latex] } .

Question 7.7.3: Solving an Equation Involving Inverse Trigonometric Function...

Precalculus

Solving an Equation Involving Inverse Trigonometric Functions

Solve $\cos^{-1} x = \sin^{-1} \frac{1}{2}$ .

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Let $\sin^{-1} \frac{1}{2} = u$ . Then $\sin u = \frac{1}{2}$ , and for u in quadrant I we have the following.

$\cos^{-1} x = \sin^{-1} \frac{1}{2}$ Original equation

$\cos^{-1} x = u$ Substitute.

$\cos u = x$ Alternative form

Sketch a triangle and label it using the facts that u is in quadrant I and $\sin u = \frac{1}{2}$ . See Figure 39. Because $x = \cos u$ , we have $x = \frac{\sqrt{3}}{2}$ . The solution set is { $\frac{\sqrt{3}}{2}$ } .