Question 7.5.5: Finding Function Values Using Definitions of the Trigonometr...
Finding Function Values Using Definitions of the Trigonometric Functions
Evaluate each expression without using a calculator.
(a) \sin (\tan^{-1} \frac{3}{2}) (b) \tan (\cos^{-1} ( -\frac{ 5}{13} ))(a) Let θ = \tan^{-1} \frac{3}{2}, and thus \tan θ = \frac{3}{2}. The inverse tangent function yields values only in quadrants I and IV, and because \frac{3}{2} is positive, θ is in quadrant I. Sketch θ in quadrant I, and label a triangle, as shown in Figure 23 on the next page. By the Pythagorean theorem, the hypotenuse is \sqrt{13}. The value of sine is the quotient of the side opposite and the hypotenuse.
\sin (\tan^{-1}\frac{3}{2} ) = \sin θ = \frac{3}{\sqrt{13}}=\frac{3}{\sqrt{13}} •\frac{\sqrt{13}}{\sqrt{13}}= \frac{3\sqrt{13}}{13}
Rationalize the denominator.
(b) Let A = \cos^{-1}( -\frac{ 5}{13}). Then, \cos A = – \frac{ 5}{13}. Because \cos^{-1} x for a negative value of x is in quadrant II, sketch A in quadrant II. See Figure 24.
\tan (\cos^{-1} (-\frac{ 5}{13})) = \tan A = – \frac{12}{5}
