9.3 Double Half and Reduction Identities

Let's go one expression at a time and match each one to a double-angle form.

1. $\cos^2(75^\circ)-\sin^2(75^\circ)$

   This is exactly $\cos(2\theta)=\cos^2\theta-\sin^2\theta$ with $\theta=75^\circ$:

   \[ \begin{align*} \cos^2(75^\circ)-\sin^2(75^\circ) &=\cos(2\cdot 75^\circ)\\ &=\cos(150^\circ)\\ &=-\frac{\sqrt{3}}{2} \end{align*} \]

2. $2\cos^2(15^\circ)-1$

   This matches $\cos(2\theta)=2\cos^2\theta-1$ with $\theta=15^\circ$:

   \[ \begin{align*} 2\cos^2(15^\circ)-1 &=\cos(2\cdot 15^\circ)\\ &=\cos(30^\circ)\\ &=\frac{\sqrt{3}}{2} \end{align*} \]

3. $1-2\sin^2(22.5^\circ)$

   This matches $\cos(2\theta)=1-2\sin^2\theta$ with $\theta=22.5^\circ$:

   \[ \begin{align*} 1-2\sin^2(22.5^\circ) &=\cos(2\cdot 22.5^\circ)\\ &=\cos(45^\circ)\\ &=\frac{\sqrt{2}}{2} \end{align*} \]

4. $2\sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right)$

   This matches $\sin(2\theta)=2\sin\theta\cos\theta$ with $\theta=\frac{\pi}{8}$:

   \[ \begin{align*} 2\sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right) &=\sin\left(2\cdot\frac{\pi}{8}\right)\\ &=\sin\left(\frac{\pi}{4}\right)\\ &=\frac{\sqrt{2}}{2} \end{align*} \]

We are given $\cos\theta=-\frac{4}{5}$ and $\theta$ is in quadrant II.

If $\theta$ is in quadrant II, then sine is positive. Build the reference triangle:

• adjacent $=-4$
• hypotenuse $=5$
• opposite $=\sqrt{5^2-4^2}=3$

So

$$ \sin\theta=\frac{3}{5}. $$

Now apply the double-angle identity for sine:

So the exact value is

$$ -\frac{24}{25}. $$

We are given

$$ \cos\theta=\frac{140}{149},\qquad \frac{3\pi}{2}<\theta<2\pi. $$

That interval puts $\theta$ in quadrant IV, so $\sin\theta<0$.

Start by finding $\sin\theta$:

Now compute each requested value.

1. $\sin(2\theta)$

   \[ \begin{align*} \sin(2\theta) &=2\sin\theta\cos\theta\\ &=2\left(-\frac{51}{149}\right)\left(\frac{140}{149}\right)\\ &=-\frac{14280}{22201} \end{align*} \]

2. $\cos(2\theta)$

   \[ \begin{align*} \cos(2\theta) &=\cos^2\theta-\sin^2\theta\\ &=\left(\frac{140}{149}\right)^2-\left(-\frac{51}{149}\right)^2\\ &=\frac{19600-2601}{22201}\\ &=\frac{16999}{22201} \end{align*} \]

3. $\tan(\theta)$

   \[ \begin{align*} \tan\theta &=\frac{\sin\theta}{\cos\theta}\\ &=\frac{-\frac{51}{149}}{\frac{140}{149}}\\ &=-\frac{51}{140} \end{align*} \]

For each one, use a half-angle identity and then simplify carefully.

1. $4\sin(-22.5^\circ)$

   Use odd symmetry first: $\sin(-\alpha)=-\sin(\alpha)$.

   \[ \begin{align*} 4\sin(-22.5^\circ) &=-4\sin(22.5^\circ) \end{align*} \]

   Now use $\sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}$ with $\theta=45^\circ$. Since $22.5^\circ$ is in quadrant I, take the positive root:

   \[ \begin{align*} \sin(22.5^\circ) &=\sqrt{\frac{1-\cos(45^\circ)}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt2}{2}}{2}}\\ &=\sqrt{\frac{2-\sqrt2}{4}}\\ &=\frac{\sqrt{2-\sqrt2}}{2} \end{align*} \]

   Therefore,

   \[ \begin{align*} 4\sin(-22.5^\circ) &=-4\cdot\frac{\sqrt{2-\sqrt2}}{2}\\ &=-2\sqrt{2-\sqrt2} \end{align*} \]

2. $3\sin\left(\frac{\pi}{8}\right)$

   Here $\frac{\pi}{8}=\frac{1}{2}\cdot\frac{\pi}{4}$, so use $\theta=\frac{\pi}{4}$:

   \[ \begin{align*} \sin\left(\frac{\pi}{8}\right) &=\sqrt{\frac{1-\cos\left(\frac{\pi}{4}\right)}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt2}{2}}{2}}\\ &=\frac{\sqrt{2-\sqrt2}}{2} \end{align*} \]

   So

   \[ \begin{align*} 3\sin\left(\frac{\pi}{8}\right) &=\frac{3\sqrt{2-\sqrt2}}{2} \end{align*} \]

3. $2\cos\left(\frac{5\pi}{12}\right)$

   Notice $\frac{5\pi}{12}=\frac{1}{2}\cdot\frac{5\pi}{6}$, so use $\cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}}$ with $\theta=\frac{5\pi}{6}$. Since $\frac{5\pi}{12}$ is in quadrant I, take the positive root:

   \[ \begin{align*} \cos\left(\frac{5\pi}{12}\right) &=\sqrt{\frac{1+\cos\left(\frac{5\pi}{6}\right)}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}\\ &=\sqrt{\frac{2-\sqrt3}{4}}\\ &=\frac{\sqrt{2-\sqrt3}}{2} \end{align*} \]

   Multiply by 2:

   \[ \begin{align*} 2\cos\left(\frac{5\pi}{12}\right) &=\sqrt{2-\sqrt3} \end{align*} \]

4. $\cos(75^\circ)$

   Again use a half-angle with $75^\circ=\frac{1}{2}(150^\circ)$:

   \[ \begin{align*} \cos(75^\circ) &=\sqrt{\frac{1+\cos(150^\circ)}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}\\ &=\sqrt{\frac{2-\sqrt3}{4}}\\ &=\frac{\sqrt{2-\sqrt3}}{2} \end{align*} \]

Given

$$ \sin\theta=\frac{8}{15},\qquad 0<\theta<\frac{\pi}{2}, $$

we know $\theta$ is in quadrant I, so all trig values for $\theta$ are positive. First find $\cos\theta$:

Also, since $0<\theta<\frac{\pi}{2}$, we have $0<\frac{\theta}{2}<\frac{\pi}{4}$, so $\frac{\theta}{2}$ is in quadrant I. That means we take positive roots for sine and cosine half-angle formulas.

1. $\sin\left(\frac{\theta}{2}\right)$

   \[ \begin{align*} \sin\left(\frac{\theta}{2}\right) &=\sqrt{\frac{1-\cos\theta}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt{161}}{15}}{2}}\\ &=\sqrt{\frac{15-\sqrt{161}}{30}} \end{align*} \]

2. $\cos\left(\frac{\theta}{2}\right)$

   \[ \begin{align*} \cos\left(\frac{\theta}{2}\right) &=\sqrt{\frac{1+\cos\theta}{2}}\\ &=\sqrt{\frac{1+\frac{\sqrt{161}}{15}}{2}}\\ &=\sqrt{\frac{15+\sqrt{161}}{30}} \end{align*} \]

3. $\tan\left(\frac{\theta}{2}\right)$

   Use $\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{\sin\theta}$:

   \[ \begin{align*} \tan\left(\frac{\theta}{2}\right) &=\frac{1-\frac{\sqrt{161}}{15}}{\frac{8}{15}}\\ &=\frac{15-\sqrt{161}}{8} \end{align*} \]

Given

$$ \cos\theta=-\frac{12}{13},\qquad \pi<\theta<\frac{3\pi}{2}, $$

$\theta$ is in quadrant III, so $\sin\theta<0$.

From a reference triangle,

Now determine the quadrant for the half-angle: if $\pi<\theta<\frac{3\pi}{2}$, then

$$ \frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}, $$

so $\frac{\theta}{2}$ is in quadrant II. Therefore:

• $\sin\left(\frac{\theta}{2}\right)>0$
• $\cos\left(\frac{\theta}{2}\right)<0$
• $\tan\left(\frac{\theta}{2}\right)<0$

1. $\sin\left(\frac{\theta}{2}\right)$

   \[ \begin{align*} \sin\left(\frac{\theta}{2}\right) &=+\sqrt{\frac{1-\cos\theta}{2}}\\ &=\sqrt{\frac{1-\left(-\frac{12}{13}\right)}{2}}\\ &=\sqrt{\frac{\frac{25}{13}}{2}}\\ &=\sqrt{\frac{25}{26}}\\ &=\frac{5\sqrt{26}}{26} \end{align*} \]

2. $\cos\left(\frac{\theta}{2}\right)$

   \[ \begin{align*} \cos\left(\frac{\theta}{2}\right) &=-\sqrt{\frac{1+\cos\theta}{2}}\\ &=-\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}\\ &=-\sqrt{\frac{\frac{1}{13}}{2}}\\ &=-\sqrt{\frac{1}{26}}\\ &=-\frac{\sqrt{26}}{26} \end{align*} \]

3. $\tan\left(\frac{\theta}{2}\right)$

   Use $\tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{\sin\theta}$:

   \[ \begin{align*} \tan\left(\frac{\theta}{2}\right) &=\frac{1-\left(-\frac{12}{13}\right)}{-\frac{5}{13}}\\ &=\frac{\frac{25}{13}}{-\frac{5}{13}}\\ &=-5 \end{align*} \]