Question 7.7.4: Solving an Inverse Trigonometric Equation Using an Identity ...

Isolate one inverse function on one side of the equation.

\arcsin x – \arccos x = \frac{π}{6}          Original equation

\arcsin x = \arccos x + \frac{π}{6}        Add arccos x. (1)

x = \sin (\arccos x +\frac{π}{6})         Definition of arcsine

Let u = \arccos x. The arccosine function yields angles in quadrants I and II, so 0 ≤ u ≤ π by definition.

x = \sin (u + \frac{π}{6})                                    Substitute.

x = \sin u \cos\frac{π}{6} + \cos u \sin \frac{π}{6}           Sine sum identity (2)

Use equation (1) and the definition of the arcsine function.

-\frac{π}{2} ≤ \arccos x +\frac{π}{6} ≤ \frac{π}{2}            Range of arcsine is [ – \frac{π}{2} , \frac{π}{2} ] .

– \frac{2π}{3} ≤\arccos x ≤ \frac{π}{3}                    Subtract \frac{π}{6} from each part.

Because both 0 ≤ \arccos x ≤ π and – \frac{2π}{3}≤ \arccos x ≤ \frac{π}{3}, the intersection yields 0 ≤ \arccos x ≤ \frac{π}{3} . This places u in quadrant I, and we can sketch the triangle in Figure 40. From this triangle we find that \sin u =\sqrt{1 – x²}. Now substitute into equation (2) using \sin u =\sqrt{1 – x²} , \sin \frac{π}{6} =\frac{ 1}{ 2} , \cos \frac{π}{6} = \frac{\sqrt{3}}{ 2} , and \cos u = x.

x = \sin u \cos \frac{π}{6}+ \cos u \sin \frac{π}{6}        (2)

x = ( \sqrt{1 – x²}) \frac{\sqrt{3}}{ 2} + x • \frac{1}{2}                 Substitute.

2x = ( \sqrt{1 – x²} )\sqrt{3} + x                   Multiply by 2.

x = ( \sqrt{3}) \sqrt{1 – x²}                                Subtract x; commutative property

x² = 3(1 – x²)                                     Square each side; (ab)² = a² b²

↑

x² = 3 – 3x²                                        Distributive property

x² = \frac{3}{4}                                              Add 3x². Divide by 4.

x = \frac{\sqrt{3}}{ 4}                                           Take the square root on each side.

↑

x = \frac{\sqrt{3}}{ 2}                                         Quotient rule: \sqrt[n]{\frac{a}{b} } =\frac{\sqrt[n]{a} }{\sqrt[n]{b} }

CHECK        A check is necessary because we squared each side when solving the equation.

\arcsin x -\arccos x = \frac{π}{6}               Original equation

\arcsin \frac{\sqrt{3}}{ 2} – \arccos \frac{\sqrt{3}}{ 2}≟ \frac{π}{6}            Let x = \frac{\sqrt{3}}{ 2} .

\frac{π}{3}- \frac{π}{6} ≟\frac{π}{6}                                     Substitute inverse values.

\frac{π}{6} = \frac{π}{6}         ✓                                  True

The solution set is { \frac{\sqrt{3}}{ 2}} .