Section 6.3: Double and Half Angle Identities

We sketch the terminal side of $\theta$ below. We think of a reference triangle formed with $x = -3$ and $y = 4$, so the hypotenuse is $$(-3)^2+4^2 = r^2$$ $$9 + 16 = r^2$$ $$r^2 = 25$$ $$r = 5$$ Hence, $\cos(\theta) = -\frac{3}{5}$ and $\sin(\theta) = \frac{4}{5}$.

Our cosine double angle identities give us three different formulas to choose from to find $\cos(2\theta)$. Using the first formula, we get: $\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta) = \left(-\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = -\frac{7}{25}$. For $\sin(2\theta)$, we get $\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25}$.

Since both cosine and sine of $2\theta$ are negative, the terminal side of $2\theta$, when plotted in standard position, lies in Quadrant III.

We can draw a reference triangle with this tangent and use it to find the sine and cosine of $\theta$. Not that since $\frac{\pi}{2}<\theta<\pi$, the terminal side of $\theta$ falls in QII.

The hypotenuse of this triangle is $$r^2 = (-4)^2+3^2$$ $$r^2 = 16 +9$$ $$r^2 = 25$$ $$r = 5$$ so $\sin(\theta) = \frac{3}{5}$ and $\cos(\theta) = -\frac{4}{5}$. Now we have everything we need to answer the rest of the question.

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ $$\sin(2\theta=2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)$$ $$\sin(2\theta=-\frac{24}{25}$$

$$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$$ $$\cos(2\theta) = \left(-\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2$$ $$\cos(2\theta) = \frac{16}{25}-\frac{9}{25}$$ $$\cos(2\theta) = \frac{7}{25}$$

$$\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}$$ $$\tan(2\theta) = \frac{-24/25}{7/25}$$ $$\tan(2\theta) = \frac{-24}{7}$$

In order to wrap our heads around what's going on here, we'll first rename the arctrig portion of the expression. Let $x = \arccos\left(\frac16\right)$. That tells us a few things:

1. The angle $x$ must be between $0$ and $\pi$.
2. $\cos(x) = \frac16$
3. Since the cosine is positive and $x$ is between $0$ and $\pi$, we can narrow our range to conclude that $x$ falls in quadrant I, that is, $0 \leq x \leq \frac{\pi}{2}$.

Now, let's return to our original question. We want to find $\sin\left(2\textcolor{blue}{\arccos\left(\frac16\right)}\right)$, and since we let $\textcolor{blue}{x = \arccos\left(\frac16\right)}$, we can rewrite our question as $\sin\left(2\textcolor{blue}{x}\right)$. According to our shiny new double angle identities, $\sin(2x) = 2\sin(x)\cos(x)$. We know that $\cos(x) = \frac16$, but we still need to find the value of $\sin(x)$. Since $x$ falls in quadrant I, we at least know that this sine value should be positive. We can draw a triangle or use the Pythagorean Identity to find the value of $\sin(x)$. $$\cos^2(x) + \sin^2(x) = 1$$ $$\left(\frac16\right)^2 + \sin^2(x) = 1$$ $$\frac{1}{36} + \sin^2(x) = \frac{36}{36}$$ $$\sin^2(x) = \frac{35}{36}$$ $$\sin(x) = \pm \sqrt{\frac{35}{36}}$$ $$\sin(x) = + \frac{\sqrt{35}}{6} \text{ (positive since x falls in QI)}$$ It's time to wrap this up! $$\sin\left(2\textcolor{blue}{\arccos\left(\frac16\right)}\right)$$ $$= \sin\left(2\textcolor{blue}{x}\right)$$ $$=2\sin(x)\cos(x)$$ $$=2\left(\frac{\sqrt{35}}{6}\right)\left(\frac16\right)$$ $$=\frac{2\sqrt{35}}{36}$$

$$\begin{array}{ll} \frac{2\tan(\theta)}{1+\tan^2(\theta)} &\\ &\\ =\frac{2\tan(\theta)}{\sec^2(\theta)} &\text{Pythagorean Identity}\\ &\\ =\frac{2\sin(\theta)/\cos(\theta)}{1/\cos^2(\theta)} &\text{Quotient & Reciprocal Identities}\\ &\\ =\frac{2\sin(\theta)}{\cos(\theta)}\cdot \frac{\cos^2(\theta)}{1} &\text{Fraction division}\\ &\\ =\frac{2\sin(\theta)}{\cancel{\cos(\theta)}}\cdot \frac{\cos^\cancel{2}(\theta)}{1} &\text{Simplify}\\ &\\ =2\sin(\theta)\cos(\theta)&\\ &\\ =\sin(2\theta) &\text{Sine double angle identity}\\ \end{array}$$

$$\begin{array}{ll} \frac{\textcolor{blue}{\sin(2x)}}{1+\textcolor{red}{\cos(2x)}}\tan^2(x) &\\ &\\ =\frac{\textcolor{blue}{2\sin(x)\cos(x)}}{1+\textcolor{red}{2\cos^2(x) - 1}}\tan^2(x) &\text{Double angle identities}\\ &\\ =\frac{2\sin(x)\cos(x)}{2\cos^2(x)}\tan^2(x) &\text{Simplify}\\ &\\ =\frac{\sin(x)}{\cos(x)}\tan^2(x) &\text{Simplify}\\ &\\ =\tan(x)\tan^2(x) &\text{Quotient Identity}\\ &\\ =\tan^3(x) &\text{Simplify}\\ \end{array}$$

We start with a clever rewrite, and then proceed with actual identities.

$$\begin{array}{ll} \cos(3x) = \cos(2x+x) &\\ &\\ =\cos(2x)\cos(x)-\sin(2x)\sin(x) &\text{cosine of a sum identity}\\ &\\ =(2\cos^2(x)-1)\cos(x)-2\sin(x)\cos(x)\sin(x) &\text{double angle identities}\\ &\\ =2\cos^3(x)-\cos(x)-2\sin^2(x)\cos(x) &\text{simplify}\\ &\\ =2\cos^3(x) - \cos(x) - 2(1-\cos^2(x))\cos(x) &\text{Pythagorean identity}\\ &\\ =2\cos^3(x)-\cos(x)-2\cos(x) + 2\cos^3(x) &\text{distribute}\\ &\\ =4\cos^3(x) - 3\cos(x)&\\ \end{array}$$

We begin with a straightforward application of the power reduction formulas:

\[ \begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \left( \dfrac{1 - \cos(2\theta)}{2} \right) \left( \dfrac{1 + \cos(2\theta)}{2} \right) \\ & = & \dfrac{1}{4}\left(1 - \cos^{2}(2\theta)\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ \end{array} \]

Next, we apply the power reduction formula to $\cos^{2}(2\theta)$ to finish the reduction

\[ \begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ & = & \dfrac{1}{4} - \dfrac{1}{4} \left(\dfrac{1 + \cos(2(2\theta))}{2}\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ & = & \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ \end{array} \]

Again, we'll start with a clever rewrite, then apply the reduction formula for cosine twice.

\[\begin{array}{rclr} \cos^4(x) & = & (\cos^2(x))^2&\\ &&&\\ &=&\left(\frac{1+\cos(2x)}{2}\right)^2 &\text{Reduction Formula}\\ &&&\\ &=&\frac{1}{4}+\frac{1}{2}\cos(2x)+\frac{1}{4}\cos^2(2x) &\text{Expand}\\ \end{array}\]

We still have a power of a cosine, so we need to apply a reduction formula again. Here, we want to think of the $\theta$ as being 2x, so $\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}$ will be $\cos^2(2x) = \frac{1+\cos(2\cdot 2x)}{2}$ when we apply the reduction formula.

\[\begin{array}{rclr} &=&\frac{1}{4}+\frac{1}{2}\cos(2x)+\frac{1}{4}\left(\frac{1+\cos(4x)}{2}\right)&\text{reduction formula}\\ &&&\\ &=&\frac{1}{4}+\frac{1}{2}\cos(2x)+\frac{1}{8}+\frac{1}{8}\cos(4x)&\text{distribute}\\ &&&\\ &=&\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)&\text{Combine like terms}\\ \end{array}\]

To use the half angle formula, we note that $\frac{\pi}{12}= \frac{\frac{\pi}{6}}{2}$ and since $\frac{\pi}{12}$ is a Quadrant I angle, its cosine is positive. Thus we have

\[ \begin{array}{rcl} \cos\left(\frac{\pi}{12}\right) & = & + \sqrt{\dfrac{1+\cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}}\\ & = & \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}\cdot \dfrac{2}{2}} = \sqrt{\dfrac{2+\sqrt{3}}{4}} = \dfrac{\sqrt{2+\sqrt{3}}}{2}\\ \end{array}\]

Back in Section 6.2, we found $\cos\left(15^{\circ}\right) = \frac{\sqrt{6}+ \sqrt{2}}{4}$ by using the difference formula for cosine. The reader is encouraged to prove that these two expressions are equal algebraically.

If $-\pi \leq t \leq 0$, then $-\frac{\pi}{2} \leq \frac{t}{2} \leq 0$, which means $ \frac{t}{2}$ corresponds to a Quadrant IV angle. Hence, $\sin\left(\frac{t}{2}\right) < 0$, so we choose the negative root formula from our half angle identities: \[ \begin{array}{rcl} \sin\left(\dfrac{t}{2} \right) & = & -\sqrt{\dfrac{1-\cos\left(t \right)}{2}} = -\sqrt{\dfrac{1- \left(-\frac{3}{5}\right)}{2}}\\ & = & -\sqrt{\dfrac{1 + \frac{3}{5}}{2} \cdot \dfrac{5}{5}} = -\sqrt{\dfrac{8}{10}} = -\dfrac{2\sqrt{5}}{5}\\ \end{array}\]

Using the given information, we can draw the triangle shown below. Using the Pythagorean Theorem, we find the hypotenuse to be 17.

Therefore, $\sin(\alpha) = -\frac{8}{17}$ and $\cos(\alpha) = -\frac{15}{17}$.

Before we continue, we must remember that although $\alpha$ is in QIII, we need to locate $\frac{\alpha}{2}$. Since $$\pi < \alpha <\frac{3\pi}{2}$$ then $$\frac{\pi}{2}<\frac{\alpha}{2} < \frac{3\pi}{4}$$ so $\frac{\alpha}{2}$ is in QII. That means $\sin\left(\frac{\alpha}{2}\right)$ will be positive. We finish with the sine half angle identity, taking the positive root.

\[\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos(\alpha)}{2}}\] \[=\sqrt{\frac{1-(-15/17)}{2}}\] \[=\sqrt{\frac{32/17}{2}}\] \[=\sqrt{\frac{16}{17}}\] \[=\frac{4}{\sqrt{17}}\] \[=\frac{4\sqrt{17}}{17}\]

First let's rename that arctrig function so that we can think of this problems in simpler terms. If we let $x = \arccos\left(-\frac35\right)$, then the problem becomes $\sin\left(\frac{\arccos\left(-\frac35\right)}{2}\right) \Rightarrow \sin\left(\frac{x}{2}\right)$. This way, we can see we're going to need a half angle identity. Now, we know that $$\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos(x)}{2}}$$ but how can we tell whether that final answer should be positive or negative? Hang on for the ride here. We know that the range of the arccosine function is $[0, \pi]$, so our angle $x$ must be between 0 and $\pi$. It follows then that $$0\leq x \leq \pi$$ $$\frac{0}{2} \leq \frac{x}{2} \leq \frac{\pi}{2}$$ $$0 \leq \frac{x}{2} \leq \frac{\pi}{2}$$ so our $\frac{x}{2}$ angle is in quadrant I, and its sine should be positive.

Now, if $x = \arccos\left(-\frac35\right)$, then by the definition of the arccosine, $\cos(x) = -\frac35$, so our final answer is: $$\sin\left(\frac{\arccos\left(-\frac35\right)}{2}\right) = \sin\left(\frac{x}{2}\right) = +\sqrt{\frac{1-(-\frac35)}{2}}$$ If we simplify, we have $$\sqrt{\frac{1+\frac35}{2}\cdot\frac{5}{5}}$$ $$=\sqrt{\frac{5+3}{10}}$$ $$=\sqrt{\frac{8}{10}}$$ $$\sin\left(\frac{\arccos\left(-\frac35\right)}{2}\right)=\sqrt{\frac{4}{5}}$$.