Chapter 7, Problem 8CR

(a)

To determine

The exact value of $\cos \theta$, if $\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Answer to Problem 8CR

Solution:

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

Explanation of Solution

Given information:

$\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Explanation:

To find $\cos \theta$,

Since $\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$,

As $\theta$ is in quadrant $\text{III}$, $\sin \theta =-\frac{1}{3}=\frac{y}{r}$

Let $y=-1$ and $r=3$

The $\theta$ is placed in the quadrant $\text{III}$, the point $P(x,y)=(x,-1)$ is on a circle of radius $r=3$,

$x^2+y^2=r^2$.

Plugging $y=-1$ and $r=3$,

$\Rightarrow x^2+{\left(-1\right)}^2=3^2$,

$\Rightarrow x^2=9-1=8$,

$\Rightarrow x=\pm \sqrt{8}$,

Since $\sin \theta$ and $\cos \theta$ are negative in the quadrant $\text{III}$,

Thus, $\sin \theta =-\frac{1}{3}=\frac{y}{r}$ and $\cos \theta =\frac{x}{r}=\frac{-\sqrt{8}}{3}=-\frac{2\sqrt{2}}{3}$

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

(b)

To determine

The exact value of $\tan \theta$ if $\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Answer to Problem 8CR

Solution:

The exact value of $\tan \theta =\frac{1}{\sqrt{8}}$ for $\pi <\theta <\frac{3\pi }{2}$,

Explanation of Solution

Given information:

$\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Explanation:

From part(a) $\cos \theta =\frac{x}{r}=\frac{-2\sqrt{2}}{3}$

As $\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$,

Therefore, $\tan \theta =\frac{\left(-\frac{1}{3}\right)}{\left(-\frac{2\sqrt{2}}{3}\right)}$.

$=\frac{1}{2\sqrt{2}}$.

$\Rightarrow \tan \theta =\frac{\sqrt{2}}{4}$.

Therefore, the exact value of $\tan \theta =\frac{\sqrt{2}}{4}$ for $\pi <\theta <\frac{3\pi }{2}$.

(c)

To determine

To calculate: The exact value of $\sin 2\theta$ if $\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Answer to Problem 8CR

Solution:

The exact value of $\sin \left(2\theta \right)=\frac{4\sqrt{2}}{9}$.

Explanation of Solution

Given information:

$\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Formula used:

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

Calculation:

From part (a) $\cos \theta =\frac{x}{r}=\frac{-2\sqrt{2}}{3}$

Since $\sin \theta$ and $\cos \theta$ are negative in the quadrant $\text{III}$,

Thus, $\sin \theta =-\frac{1}{3}=\frac{y}{r}$ and $\cos \theta =\frac{x}{r}=\frac{-2\sqrt{2}}{3}$

By using the double angel formula, $\sin \left(2\theta \right)=2\sin \theta \cos \theta$.

$\Rightarrow \sin \left(2\theta \right)=2\left(-\frac{1}{3}\right)\left(-\frac{2\sqrt{2}}{3}\right)=\frac{4\sqrt{2}}{9}$,

Therefore, the exact value of $\sin \left(2\theta \right)=\frac{4\sqrt{2}}{9}$.

(d)

To determine

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

Answer to Problem 8CR

Solution:

The exact value of $\cos \left(2\theta \right)=\frac{7}{9}$ for $\pi <\theta <\frac{3\pi }{2}$,

Explanation of Solution

Given information:

$\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Formula used:

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

Calculation:

$\sin \theta =\frac{y}{r}=-\frac{1}{3}$,

By using double angel formula, $\cos \left(2\theta \right)=1-2{\sin }^2\theta$.

$\Rightarrow \cos \left(2\theta \right)=1-2{\left(-\frac{1}{3}\right)}^2=1-\frac{2}{9}$,

Thus, $\cos \left(2\theta \right)=\frac{7}{9}$ for $\pi <\theta <\frac{3\pi }{2}$,

Therefore, the exact value of $\cos \left(2\theta \right)=\frac{7}{9}$ for $\pi <\theta <\frac{3\pi }{2}$,

(e)

To determine

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

Answer to Problem 8CR

Solution:

The exact value of $\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{3+2\sqrt{2}}{6}}$.

Explanation of Solution

Given information:

$\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Formula used:

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

Calculation:

From part a),

$\cos \theta =-\frac{2\sqrt{2}}{3}$ and $\theta$ is in the quadrant $\text{III}$,

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

As $\theta$ is in quadrant $\text{III}$ thus $\frac{\theta }{2}$ is lies in the quadrant $\text{II}$,

Since the value of $\sin (\theta )$ in quadrant $\text{II}$ is positive thus the sign of formula is $+$,

$\Rightarrow \sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{1-\left(-\frac{2\sqrt{2}}{3}\right)}{2}}=\sqrt{\frac{1+\left(\frac{2\sqrt{2}}{3}\right)}{2}}=\sqrt{\frac{3+2\sqrt{2}}{3\cdot 2}}=\sqrt{\frac{3+2\sqrt{2}}{6}}$,

Thus, $\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{3+2\sqrt{2}}{6}}$

Therefore, the exact value of $\sin \left(\frac{\theta }{2}\right)=\sqrt{\frac{3+2\sqrt{2}}{6}}$.

(f)

To determine

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

Answer to Problem 8CR

Solution:

The exact value of $\cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{3-2\sqrt{2}}{6}}$.

Explanation of Solution

Given information:

$\sin \theta =-\frac{1}{3}$ for $\pi <\theta <\frac{3\pi }{2}$.

Formula used:

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

Calculation:

From part a),

$\cos \theta =-\frac{2\sqrt{2}}{3}$ and $\theta$ is in the quadrant $\text{III}$,

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

Since the value of $\cos (\theta )$ in quadrant $\text{II}$ is negative thus the sign of formula is $-$,

$\Rightarrow \cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{1+\left(-\frac{2\sqrt{2}}{3}\right)}{2}}=-\sqrt{\frac{1-\frac{2\sqrt{2}}{3}}{2}}=-\sqrt{\frac{3-2\sqrt{2}}{2\cdot 3}}=-\sqrt{\frac{3-2\sqrt{2}}{6}}$,

Thus, $\cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{3-2\sqrt{2}}{6}}$ for $\pi <\theta <\frac{3\pi }{2}$.

Therefore, the exact value of $\cos \left(\frac{\theta }{2}\right)=-\sqrt{\frac{3-2\sqrt{2}}{6}}$ for $\pi <\theta <\frac{3\pi }{2}$.