Question 6.6.8: Finding the Exact Value of a Composite Trigonometric Express...

a. Let θ represent the radian measure of the angle in the interval \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), with \tan \theta=\frac{2}{3}. Then because tan θ is positive, θ must be positive. We have

\theta=\tan ^{-1} \frac{2}{3} \quad \text { and } \quad 0<\theta<\frac{\pi}{2} .

Figure 85 shows θ in standard position. If (x, y) is a point on the terminal side of θ , then \tan \theta=\frac{y}{x} .

Consequently, we can choose the point with coordinates (3,2) to determine the terminal side of θ.Then x=3, y=2 and we have

\begin{gathered}\tan \theta=\frac{2}{3} \quad \text { and } \cos \theta=\frac{3}{r}, \text { where } \\r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+2^{2}}=\sqrt{9+4}=\sqrt{13} \text {. So } \\\cos \left(\tan ^{-1} \frac{2}{3}\right)=\cos \theta=\frac{3}{r}=\frac{3}{\sqrt{13}}=\frac{3 \sqrt{13}}{13} .\end{gathered}

b. Let θ represent the radian measure of the angle in [0, \pi], \text { with } \cos \theta=-\frac{1}{4} .
Because cos θ is negative, θ is in quadrant II; so

\theta=\cos ^{-1}\left(-\frac{1}{4}\right) \quad \text { and } \quad \frac{\pi}{2}<\theta<\pi .

Figure 86 shows θ in standard position. If (x, y) is a point on the terminal side of θ and r is the distance between (x, y) and the origin, then \sin \theta=\frac{y}{r} . We choose the point with coordinates (-1, y), a distance of four units from the origin, on the terminal side of θ. Then

\begin{aligned}&r=\sqrt{x^{2}+y^{2}}=\sqrt{(-1)^{2}+y^{2}} \text { or } r^{2}=1+y^{2}\\&4^{2}=1+y^{2} \quad \text { Replace } r \text { with } 4 \text {. }\\&15=y^{2} \quad \text { Simplify. }\\&\sqrt{15}=y \quad y \text { is positive. }\end{aligned}

Thus,

\sin \left[\cos ^{-1}\left(-\frac{1}{4}\right)\right]=\sin \theta=\frac{y}{r}=\frac{\sqrt{15}}{4}.