Question

1. evaluate each expression. assume that all angles are in quadrant i.

a. ( \tan left( cos ^ { - 1 } \frac { 2 } { 3 }
ight) )
b. ( sin left( arctan \frac { 5 } { 4 }
ight) )
c. ( cos left( cos ^ { - 1 } left( \frac { 2 } { 3 }
ight)
ight) )

Explanation:

Step1: Let \(\theta=\cos^{-1}\frac{2}{3}\)

By the definition of inverse cosine, \(\cos\theta=\frac{2}{3}\). In a right - triangle (since \(\theta\) is in quadrant I), if the adjacent side \(x = 2\) and the hypotenuse \(r=3\), then by the Pythagorean theorem \(y=\sqrt{r^{2}-x^{2}}=\sqrt{9 - 4}=\sqrt{5}\).
And \(\tan\theta=\frac{y}{x}\).
So, \(\tan(\cos^{-1}\frac{2}{3})=\frac{\sqrt{5}}{2}\).

Step2: Let \(\alpha=\arctan\frac{5}{4}\)

By the definition of inverse tangent, \(\tan\alpha=\frac{5}{4}\). In a right - triangle (since \(\alpha\) is in quadrant I), if the opposite side \(y = 5\) and the adjacent side \(x = 4\), then the hypotenuse \(r=\sqrt{x^{2}+y^{2}}=\sqrt{16 + 25}=\sqrt{41}\).
And \(\sin\alpha=\frac{y}{r}\).
So, \(\sin(\arctan\frac{5}{4})=\frac{5}{\sqrt{41}}=\frac{5\sqrt{41}}{41}\).

Step3: Use the property of inverse cosine function

For \(y = \cos^{-1}x\), the domain of \(x\) is \([-1,1]\) and \(\cos(\cos^{-1}x)=x\) when \(x\in[-1,1]\).
Since \(\frac{2}{3}\in[-1,1]\), then \(\cos(\cos^{-1}(\frac{2}{3}))=\frac{2}{3}\).

Answer:

a. \(\frac{\sqrt{5}}{2}\)
b. \(\frac{5\sqrt{41}}{41}\)
c. \(\frac{2}{3}\)