Chapter 4, Problem 33RE

To determine

To Find:

The exact values of 6 trigonometric function of the angle $\theta$

Answer to Problem 33RE

Third $\sin \theta =\frac{4}{\sqrt{41}}=0.624,\cos ec\theta =\frac{\sqrt{41}}{4}=1.600$

Side is $\sqrt{41}$

  $\cos \theta =\frac{5}{\sqrt{41}}=0.780,sec\theta =\frac{\sqrt{41}}{5}=1.280$

  $\tan \theta =\frac{4}{5}=0.80,\cot \theta =\frac{5}{4}=1.25$

Explanation of Solution

Given information:

Side are given, perpendicular side is 4.

Base is 5.

Formula used:

Pythagoras theorem formula.

  $h=\sqrt{p^2+b^2}$

And trigonometric formula

  $\sin \theta =\frac{perpendicular}{height}$

  $\cos \theta =\frac{base}{height}$

  $\tan \theta =\frac{perpendicular}{base}$

  $\cos ec\theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

Consider the statement “ exact value of 6 trigonometic function.”

Recall that Pythagoras theorem

  $h=\sqrt{p^2+b^2}$ ,where h is height p is perpendicular , b is base.

= $\sqrt{4^2+5^2}$

= $\sqrt{16+25}$

= $\sqrt{41}$

Height, h= $\sqrt{41}$

Recall the trigonometric function

  $\sin \theta =\frac{perpendicular}{height}$

  $=\frac{4}{\sqrt{41}}$

  $=0.624$

Therefore,

  $\sin \theta =0.624$

  $\cos \theta =\frac{base}{height}$

  $=\frac{5}{\sqrt{41}}=0.780$

Therefore, $\cos \theta =0.780$

  $\tan \theta =\frac{perpendicular}{base}$

  $=\frac{4}{5}=0.80$

Therefore, $\tan \theta =0.80$

  $\cos ec\theta =\frac{1}{\sin \theta }$

  $=\frac{\sqrt{41}}{4}=1.600$

Therefore, $\cos ec\theta =1.600$

  $\sec \theta =\frac{1}{\cos \theta }$

  $=\frac{\sqrt{41}}{5}=1.280$

Therefore, $\sec \theta =1.280$

  $\cot \theta =\frac{1}{\tan \theta }$

  $=\frac{5}{4}=1.25$

So, the third side is $h=\sqrt{41}$ and trigonometric function

  $\begin{array}{l}\sin \theta =0.624\\ \cos \theta =0.780\\ \tan \theta =0.80\\ \cos ec\theta =1.600\\ \sec \theta =1.280\\ \cot \theta =1.25\end{array}$ where h is height,p is perpendicular,b is base.