Chapter 8, Problem 6CR

(a)

To determine

The exact value of $\sin \theta$, if $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Answer to Problem 6CR

Solution:

The exact value of $\sin \theta =-\frac{2\sqrt{5}}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

Explanation of Solution

Given information:

  $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

As $\theta$ is in quadrant $\text{IV}$, only $\cos \theta$ and $\sec \theta$ are positive

  $\tan \theta =-2=\frac{y}{x}$

  $\Rightarrow y=-2$ and $x=1$

The $\theta$ is in the quadrant $\text{IV}$, the point $P(x,y)=(1,-2)$ is on circle of radius $r$

The equation of circle is $x^2+y^2=r^2$

Substitute $y=-2$ and $x=1$ in the above equation

  $\Rightarrow 1^2+{\left(-2\right)}^2=r^2$ ,

  $\Rightarrow r^2=1+4=5$ ,

  $\Rightarrow r=\sqrt{5}$ ,

Since $\sin \theta$ is negative in quadrant $\text{IV}$

Thus, $\sin \theta =\frac{y}{r}=\frac{-2}{\sqrt{5}}=\frac{-2\sqrt{5}}{5}$

Therefore, $\sin \theta =\frac{-2\sqrt{5}}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

(b)

To determine

The exact value of $\cos \theta$, if $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Answer to Problem 6CR

Solution:

The exact value of $\cos \theta =\frac{\sqrt{5}}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

Explanation of Solution

Given information:

  $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

From part(a), $x=1,y=-2$ and $r=\sqrt{5}$

As $\cos \theta =\frac{x}{r}$ ,

Substitute the value $x=1$ and $r=\sqrt{5}$

  $\Rightarrow \cos \theta =\frac{\sqrt{5}}{5}$

Therefore, the exact value of $\cos \theta =\frac{\sqrt{5}}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

(c)

To determine

To calculate: The exact value of $\sin 2\theta$, if $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Answer to Problem 6CR

Solution:

The exact value of $\sin \left(2\theta \right)=-\frac{4}{5}$

Explanation of Solution

Given information:

  $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Formula used:

The double angel formula: $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

Calculation:

From part (a) and part (b),

  $\sin \theta =-\frac{2\sqrt{5}}{5}$ and $\cos \theta =\frac{\sqrt{5}}{5}$

Substitute these values in the double angle formula

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$ .

  $\Rightarrow \sin \left(2\theta \right)=2\left(-\frac{2\sqrt{5}}{5}\right)\left(\frac{\sqrt{5}}{5}\right)=-\frac{4}{5}$ ,

Therefore, the exact value of $\sin \left(2\theta \right)=-\frac{4}{5}$

(d)

To determine

To calculate: The exact value of $\cos \left(2\theta \right)$, if $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Answer to Problem 6CR

Solution:

The exact value of $\cos \left(2\theta \right)=-\frac{3}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

Explanation of Solution

Given information:

  $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Formula used:

The double angel formula: $\cos \left(2\theta \right)=1-2{\sin }^2\theta$

Calculation:

By part (a), $\sin \theta =-\frac{2\sqrt{5}}{5}$ .

Substitute this value in the double angle formula $\cos \left(2\theta \right)=1-2{\sin }^2\theta$

  $\Rightarrow \cos \left(2\theta \right)=1-2{\left(-\frac{2\sqrt{5}}{5}\right)}^2=1-2\left(\frac{4}{5}\right)$ ,

  $\Rightarrow \cos \left(2\theta \right)=1-\frac{8}{5}=-\frac{3}{5}$

Therefore, $\cos \left(2\theta \right)=-\frac{3}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

Thus the exact value of $\cos \left(2\theta \right)=-\frac{3}{5}$ for $\frac{3\pi }{2}<\theta <2\pi$

(e)

To determine

To calculate: The exact value of $\sin \left(\frac{1}{2}\theta \right)$ , if $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Answer to Problem 6CR

Solution:

The exact value of $\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{5-\sqrt{5}}{10}}$

Explanation of Solution

Given information:

  $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Formula used:

The half angel formula: $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$

Calculation:

From part b),

  $\cos \theta =\frac{\sqrt{5}}{5}$ and $\theta$ is in quadrant $\text{IV}$

By using the half angel formula: $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$

As $\frac{3\pi }{2}<\theta <2\pi$

Divide the inequality by $2$

  $\Rightarrow \frac{3\pi }{4}<\frac{\theta }{2}<\pi$ .

That is, half angle lies in quadrant $\text{II}$

As value of sine function is positive in quadrant $\text{II}$, the formula of half angle is positive

  $\Rightarrow \sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\frac{\sqrt{5}}{5}}{2}}=\sqrt{\frac{\frac{5-\sqrt{5}}{5}}{2}}=\sqrt{\frac{5-\sqrt{5}}{10}}$ ,

Therefore, 00 $\sin \left(\frac{1}{2}\theta \right)=\sqrt{\frac{5-\sqrt{5}}{10}}$

Thus, the exact value of $\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{5-\sqrt{5}}{10}}$

(f)

To determine

To calculate: The exact value of $\cos \left(\frac{1}{2}\theta \right)$ , if $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Answer to Problem 6CR

Solution:

The exact value of $\cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{5+\sqrt{5}}{10}}$

Explanation of Solution

Given information:

  $\tan \theta =-2$ for $\frac{3\pi }{2}<\theta <2\pi$

Formula used:

The half angel formula: $\cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \theta }{2}}$

Calculation:

From part b),

  $\cos \theta =\frac{\sqrt{5}}{5}$ and $\theta$ is in quadrant $\text{IV}$

As $\frac{3\pi }{2}<\theta <2\pi$ .

Divide the inequality by $2$

  $\Rightarrow \frac{3\pi }{4}<\frac{\theta }{2}<\pi$ .

That is, half angle lies in quadrant0 $\text{II}$

As value of cosine function is negative in quadrant $\text{II}$, the formula of the half angle is negative.

By using the half angel formula:00 $\cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \theta }{2}}$

0 $\Rightarrow \cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{1+\frac{\sqrt{5}}{5}}{2}}=-\sqrt{\frac{5+\sqrt{5}}{10}}$ ,

Therefore, $\cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{5+\sqrt{5}}{10}}$ for $\frac{3\pi }{2}<\theta <2\pi$

Thus, the exact value of $\cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{5+\sqrt{5}}{10}}$ For0 $\frac{3\pi }{2}<\theta <2\pi$