Chapter 13.4, Problem 1CYU

a)

To determine

To Find : The value of $\sin 2\theta$

Answer to Problem 1CYU

  $\frac{\sqrt{15}}{8}$

Explanation of Solution

Given Information:

We know that, the given expression is the value of $\sin \theta =\frac{1}{4},0^{\circ }<\theta <{90}^{\circ }$

Formula used: $\sin 2\theta =2\sin \theta \cos \theta$

Calculation:

  $\begin{array}{l}\sin \theta =\frac{1}{4}\\ \cos \theta =\frac{\sqrt{15}}{4}\\ \sin 2\theta =2\sin \theta \cos \theta \\ \sin 2\theta =2\times \frac{1}{4}\times \frac{\sqrt{15}}{4}=\frac{\sqrt{15}}{8}\end{array}$

  

b)

To determine

To Find: The value of $\cos 2\theta$

Answer to Problem 1CYU

  $\frac{7}{8}$

Explanation of Solution

Given Information:

We know that, the given expression is the value of $\sin \theta =\frac{1}{4},0^{\circ }<\theta <{90}^{\circ }$

Formula used:

  $\cos 2\theta =2{\cos }^2\theta -1$

Calculation:

  $\begin{array}{l}\cos 2\theta =2{\cos }^2\theta -1\\ =2\times {\left(\frac{\sqrt{15}}{4}\right)}^2-1\\ =\frac{30}{16}-1\\ \cos 2\theta =\frac{7}{8}\end{array}$

c)

To determine

To Find: The value of $\sin \frac{\theta }{2}$

Answer to Problem 1CYU

  $\frac{\sqrt{4-\sqrt{5}}}{2\sqrt{2}}$

Explanation of Solution

Given Information:

We know that, the given expression is the value of $\sin \theta =\frac{1}{4},0^{\circ }<\theta <{90}^{\circ }$

Formula used:

  $\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}}$

Calculation:

  $\begin{array}{l}\sin \frac{\theta }{2}=\sqrt{\frac{1-\frac{\sqrt{15}}{4}}{2}}\\ \sin \frac{\theta }{2}=\frac{\sqrt{4-\sqrt{5}}}{2\sqrt{2}}\end{array}$

d)

To determine

To Find: The value of $\cos \frac{\theta }{2}$

Answer to Problem 1CYU

  $\frac{\sqrt{8-2\sqrt{5}}}{4}$

Explanation of Solution

Given Information:

The given expression is the value of $\sin \theta =\frac{1}{4},0^{\circ }<\theta <{90}^{\circ }$

Formula used: $\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$

Calculation:

  $\begin{array}{l}\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\\ \frac{1}{4}=2\times \frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}}\times \cos \frac{\theta }{2}\\ \cos \frac{\theta }{2}=\frac{\sqrt{8-2\sqrt{5}}}{4}\end{array}$