Chapter 7.6, Problem 11AYU

In Problems $9-20,$ use the information given about the angle $\theta ,0\le \theta <2\pi ,$ to find the exact value of:

$(a)\sin (2\theta )(b)\cos (2\theta )(c)\sin \frac{\theta }{2}(d)\cos \frac{\theta }{2}$

$\cos \theta =-\frac{\sqrt{6}}{3},\frac{\pi }{2}<\theta <\pi$

To determine

To calculate: The exact value of $\sin \left(2\theta \right)$ if $\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Answer to Problem 11AYU

Solution:

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

Explanation of Solution

Given information:

$\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Formula used:

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

Calculation:

To find $\sin \theta$,

$\cos \theta =\frac{x}{r}=-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

As $\theta$ is in quadrant $\text{II}$,

let $x=-\sqrt{6}$ and $r=3$.

The $\theta$ is placed in the quadrant $\text{II}$, the point $P(x,y)=(-\sqrt{6},y)$ is on a circle of radius $3$,

$x^2+y^2=9$.

Substituting $x=-\sqrt{6}$,

$\Rightarrow {\left(-\sqrt{6}\right)}^2+y^2=9$

$\Rightarrow y^2=9-6=3$

$\Rightarrow y=\pm \sqrt{3}$

Since, $\theta$ is placed in the quadrant $\text{II}$.

$\Rightarrow y=\sqrt{3}$

Thus $\sin \theta =\frac{y}{r}=\frac{\sqrt{3}}{3}$.

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

$\Rightarrow \sin \left(2\theta \right)=2\left(\frac{\sqrt{3}}{3}\right)\left(-\frac{\sqrt{6}}{3}\right)=-\frac{2\sqrt{18}}{9}=-\frac{2\sqrt{2}}{3}$

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

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

To determine

To calculate: The exact value of $\cos \left(2\theta \right)$ if $\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Answer to Problem 11AYU

Solution:

The exact value of $\cos \left(2\theta \right)=\frac{1}{3}$.

Explanation of Solution

Given information:

$\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Formula used:

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

Calculation:

From part a),

$\sin \theta =\frac{y}{r}=\frac{\sqrt{3}}{3}$.

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

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

Thus, $\cos \left(2\theta \right)=\frac{1}{3}$.

Therefore, the exact value of $\cos \left(2\theta \right)=\frac{1}{3}$.

To determine

To calculate: The exact value of $\sin \left(\frac{\theta }{2}\right)$ for $0\le \theta <2\pi$ if $\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Answer to Problem 11AYU

Solution:

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

Explanation of Solution

Given information:

$\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Formula used:

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

Calculation:

Since, $\cos \theta =-\frac{\sqrt{6}}{3}$ and $\theta$ is in quadrant $\text{II}$.

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

As $\theta$ is in quadrant $\text{II}$ thus $\frac{\theta }{2}$ is lies in quadrant $\text{I}$.

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

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

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

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

To determine

To calculate: The exact value of $\cos \left(\frac{\theta }{2}\right)$ if $\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Answer to Problem 11AYU

Solution:

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

Explanation of Solution

Given information:

$\cos \theta =-\frac{\sqrt{6}}{3}$ for $\frac{\pi }{2}<\theta <\pi$.

Formula used:

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

Calculation:

Since, $\cos \theta =-\frac{\sqrt{6}}{3}$ and $\theta$ is in quadrant $\text{II}$,

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

As $\theta$ is in quadrant $\text{II}$ thus $\frac{\theta }{2}$ is lies in quadrant $\text{I}$.

Since the value of $\cos (\theta )$ in quadrant $\text{I}$ is positive thus the sign of formula is $+$.

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

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

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