Chapter 7.2, Problem 63AYU

To determine

To express: The trigonometric expression $\cos \left(\cos e{c}^{-1}u\right)$ as an algebraic expression in u .

Answer to Problem 63AYU

The expression is $\frac{\sqrt{u^2-1}}{u}$ .

Explanation of Solution

Given information:

The trigonometric expression $\cos \left(\cos e{c}^{-1}u\right)$ .

Formula used:

Trigonometric Identities :

Cosine of angle $\theta$ Cosecant of angle $\theta$

  $\cos \theta =\frac{Base}{Hypotenuse}$

  $\cos ec\theta =\frac{Hypotenuse}{Perpendicular}$

Pythagoras Theorem:- The hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides.

  ${\left(perpendicular\right)}^2+{\left(base\right)}^2={\left(hypotenuse\right)}^2$

Calculation:

Consider ,

  $\cos \left(\cos e{c}^{-1}u\right)$

Let

  $\begin{array}{l}\theta =\cos e{c}^{-1}u\\ \cos ec\theta =u\end{array}$ (i)

where $y=\csc x$ is between $y\le -1$ ; and $-\infty <u<\infty$ .

Now,

  $\cos \left(\cos e{c}^{-1}u\right)=\cos \theta$ (ii)

Using equation (i)

  $\begin{array}{l}=\cos ec\theta =\frac{u}{1}\\ =\frac{Hypotenuse}{Perpendicular}=\frac{u}{1}\end{array}$

  $\begin{array}{l}=P^2+B^2=H^2\\ ={(}^1+B^2={\left(u\right)}^2\\ =1+B^2=u^2\\ =\sqrt{u^2-1}=B\end{array}$

  $=\cos \theta =\frac{B}{H}=\frac{\sqrt{u^2-1}}{u}$

Hence , the algebraic expression is $\frac{\sqrt{u^2-1}}{u}$ .