Chapter 4.3, Problem 37E

(a)

To determine

Value of $\sin {30}^0$ .

Answer to Problem 37E

The value of $\sin {30}^0=\frac{1}{2}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {60}^0=\frac{\sqrt{3}}{2}\\ \cos {60}^0=\frac{1}{2}\end{array}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\sin {30}^0$ using $\sin {60}^0=\frac{\sqrt{3}}{2},\cos {60}^0=\frac{1}{2}$

We have

  $\sin {30}^0=\cos {(90-30)}^0=\cos {60}^0=\frac{1}{2}$

Value of $\sin {30}^0=\frac{1}{2}$

Conclusion:

Thus,the value of $\sin {30}^0$ is $\frac{1}{2}$

(b)

To determine

Value of $\cos {30}^0$ .

Answer to Problem 37E

The value of $\cos {30}^0=\frac{\sqrt{3}}{2}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {60}^0=\frac{\sqrt{3}}{2}\\ \cos {60}^0=\frac{1}{2}\end{array}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\cos {30}^0$ using $\sin {60}^0=\frac{\sqrt{3}}{2},\cos {60}^0=\frac{1}{2}$

We have

  $\cos {30}^0=\sin {(90-30)}^0=\sin {60}^0=\frac{\sqrt{3}}{2}$

Value of $\cos {30}^0=\frac{\sqrt{3}}{2}$

Conclusion:

Thus,the value of $\sin {30}^0$ is $\frac{\sqrt{3}}{2}$

(c)

To determine

Value of $\tan {60}^0$ .

Answer to Problem 37E

The value of $\tan {60}^0=\sqrt{3}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {60}^0=\frac{\sqrt{3}}{2}\\ \cos {60}^0=\frac{1}{2}\end{array}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\tan {60}^0$ using $\sin {60}^0=\frac{\sqrt{3}}{2},\cos {60}^0=\frac{1}{2}$

We have

  $\begin{array}{l}tan\theta =\frac{\sin \theta }{\cos \theta }\\ tan60=\frac{\sin 60}{\cos 60}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\end{array}$

Value of $\tan {60}^0=\sqrt{3}$

Conclusion:

Thus,the value of $\tan {60}^0$ is $\sqrt{3}$

(d)

To determine

Value of $\cot {60}^0$ .

Answer to Problem 37E

The value of $\cot {60}^0=\frac{1}{\sqrt{3}}$

Explanation of Solution

Given info:

  $\begin{array}{l}\sin {60}^0=\frac{\sqrt{3}}{2}\\ \cos {60}^0=\frac{1}{2}\end{array}$

Formula used:

We have the results

  $\begin{array}{l}\sin \theta =\cos (90-\theta )\\ \cos \theta =\sin (90-\theta )\\ \cos ec\theta =\sec (90-\theta )\\ \sec \theta =\cos ec(90-\theta )\\ \tan \theta =\cot (90-\theta )\\ \cot \theta =\tan (90-\theta )\\ tan\theta =\frac{\sin \theta }{\cos \theta }\\ \cot \theta =\frac{\cos \theta }{\sin \theta }\\ cosec\theta =\frac{1}{\sin \theta }\\ \sec \theta =\frac{1}{\cos \theta }\\ si{n}^2\theta +{\cos }^2\theta =1\\ {\sec }^2\theta -{\tan }^2\theta =1\\ cose{c}^2\theta -{\cot }^2\theta =1\end{array}$

Calculation:

Here we need to find $\cot {60}^0$ using $\sin {60}^0=\frac{\sqrt{3}}{2},\cos {60}^0=\frac{1}{2}$

We have

  $\begin{array}{l}\cot \theta =\frac{\cos \theta }{\sin \theta }\\ \cot 60=\frac{\cos 60}{\sin 60}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}\end{array}$

Value of $\cot {60}^0=\frac{1}{\sqrt{3}}$

Conclusion:

Thus,the value of $\cot {60}^0$ is $\frac{1}{\sqrt{3}}$