Double-angle and half-angle formulas
The sum formulas from the last lesson give formulas for twice an angle, ${2x}$. Turning these around gives formulas for half an angle, ${\dfrac{x}{2}}$, and formulas that lower powers like ${\sin^2 x}$. All of them are used often in calculus.
Double-angle formulas
Put ${B=A=x}$ into the sum formulas. For sine:
$\begin{align*}&\sin(x+x)\\&=\sin x\cos x+\cos x\sin x\end{align*}$
so ${\sin 2x=2\sin x\cos x}$. The same idea works for cosine and tangent:
${\sin 2x=2\sin x\cos x}$
${\cos 2x=\cos^2 x-\sin^2 x}$
${\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}}$
The cosine formula has two more forms. Replace ${\sin^2 x}$ by ${1-\cos^2 x}$, or replace ${\cos^2 x}$ by ${1-\sin^2 x}$:
${\cos 2x=2\cos^2 x-1}$
${\cos 2x=1-2\sin^2 x}$
Watch out: ${\sin 2x}$ is not ${2\sin x}$. For example, take ${x=\dfrac{\pi}{2}}$. Then ${\sin 2x=\sin\pi=0}$, but ${2\sin x=2}$.
Example 1: Suppose ${\sin x=\dfrac{3}{5}}$ and ${x}$ is in quadrant II. Find ${\sin 2x}$, ${\cos 2x}$, and ${\tan 2x}$.
Solution:
First find ${\cos x}$. In quadrant II it is negative:
$\cos x=-\sqrt{1-\dfrac{9}{25}}=-\dfrac{4}{5}$
Now use the formulas:
$\sin 2x=2\cdot\dfrac{3}{5}\cdot\left(-\dfrac{4}{5}\right)=-\dfrac{24}{25}$
$\begin{align*}\cos 2x&=1-2\left(\dfrac{3}{5}\right)^2\\&=1-\dfrac{18}{25}=\dfrac{7}{25}\end{align*}$
$\begin{align*}\tan 2x&=\dfrac{\sin 2x}{\cos 2x}=\dfrac{-24/25}{7/25}\\&=-\dfrac{24}{7}\end{align*}$
Example 2: Write ${\cos 3x}$ in terms of ${\cos x}$.
Solution:
Write ${3x=2x+x}$, and use the sum formula, then the double-angle formulas:
$\begin{align*}&\cos(2x+x)\\&=\cos 2x\cos x-\sin 2x\sin x\\&=(2\cos^2 x-1)\cos x-2\sin^2 x\cos x\end{align*}$
Replace ${\sin^2 x}$ by ${1-\cos^2 x}$, and collect terms:
$\begin{align*}&=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x\\&=2\cos^3 x-\cos x-2\cos x+2\cos^3 x\\&=4\cos^3 x-3\cos x\end{align*}$
Example 3: Prove the identity
${\dfrac{\sin 2x}{1+\cos 2x}=\tan x}$
Solution:
Use ${\sin 2x=2\sin x\cos x}$ on top. On the bottom, use ${\cos 2x=2\cos^2 x-1}$, so that the ${1}$'s cancel:
$\begin{align*}&\dfrac{\sin 2x}{1+\cos 2x}\\&=\dfrac{2\sin x\cos x}{1+2\cos^2 x-1}\\&=\dfrac{2\sin x\cos x}{2\cos^2 x}\\&=\dfrac{\sin x}{\cos x}=\tan x\end{align*}$
Example 4: A ball is thrown at ${64}$ ft/s, at an angle ${\theta}$ above the ground. Ignoring air resistance, it lands a distance
${R=\dfrac{64^2}{16}\sin\theta\cos\theta}$ feet
away. Which angle sends it farthest?
Solution:
Since $\sin\theta\cos\theta=\dfrac{1}{2}\sin 2\theta$, and ${\dfrac{64^2}{16}=256}$,
$R=256\cdot\dfrac{1}{2}\sin 2\theta=128\sin 2\theta$
The largest value of ${\sin 2\theta}$ is ${1}$, when ${2\theta=90^\circ}$. So the best angle is ${\theta=45^\circ}$, and the ball then goes ${128}$ ft.
Power-reducing formulas
Solve the two extra forms of ${\cos 2x}$ for ${\sin^2 x}$ and ${\cos^2 x}$. This gives formulas that replace a square by a first power of cosine:
${\sin^2 x=\dfrac{1-\cos 2x}{2}}$
${\cos^2 x=\dfrac{1+\cos 2x}{2}}$
${\tan^2 x=\dfrac{1-\cos 2x}{1+\cos 2x}}$
The graph shows the first formula at work: ${\sin^2 x}$ is a wave with period ${\pi}$, centered on ${y=\dfrac{1}{2}}$.
Example 5: Write ${\sin^4 x}$ using only first powers of cosines.
Solution:
Write ${\sin^4 x=(\sin^2 x)^2}$, and use the formula for ${\sin^2 x}$:
$\begin{align*}&\sin^4 x\\&=\left(\dfrac{1-\cos 2x}{2}\right)^2\\&=\dfrac{1-2\cos 2x+\cos^2 2x}{4}\end{align*}$
There is still a square, ${\cos^2 2x}$. Use the formula for ${\cos^2}$ again, with ${2x}$ in place of ${x}$:
${\cos^2 2x=\dfrac{1+\cos 4x}{2}}$
Put this in. The top of the fraction becomes
$\begin{align*}&1-2\cos 2x+\dfrac{1+\cos 4x}{2}\\&=\dfrac{3}{2}-2\cos 2x+\dfrac{1}{2}\cos 4x\end{align*}$
The bottom is ${4}$. Multiply the top and the bottom by ${2}$ to clear the halves:
${\sin^4 x=\dfrac{3-4\cos 2x+\cos 4x}{8}}$
Half-angle formulas
Put ${x=\dfrac{u}{2}}$ into the power-reducing formulas, and take square roots:
$\sin\dfrac{u}{2}=\pm\sqrt{\dfrac{1-\cos u}{2}}$
$\cos\dfrac{u}{2}=\pm\sqrt{\dfrac{1+\cos u}{2}}$
$\tan\dfrac{u}{2}=\dfrac{1-\cos u}{\sin u}=\dfrac{\sin u}{1+\cos u}$
The sign, ${+}$ or ${-}$, depends on the quadrant of ${\dfrac{u}{2}}$, not of ${u}$. The tangent formula has no ${\pm}$; its sign takes care of itself.
Example 6: Find the exact values.
(a) ${\sin 15^\circ}$ (b) ${\tan 22.5^\circ}$
Solution:
(a) ${15^\circ}$ is half of ${30^\circ}$, and it is in quadrant I, so use ${+}$:
$\begin{align*}\sin 15^\circ&=\sqrt{\dfrac{1-\cos 30^\circ}{2}}=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}\\&=\sqrt{\dfrac{2-\sqrt{3}}{4}}=\dfrac{\sqrt{2-\sqrt{3}}}{2}\end{align*}$
(This looks different from ${\dfrac{\sqrt{6}-\sqrt{2}}{4}}$ in the last lesson, but both are about ${0.2588}$.)
(b) ${22.5^\circ}$ is half of ${45^\circ}$:
$\begin{align*}\tan 22.5^\circ&=\dfrac{1-\cos 45^\circ}{\sin 45^\circ}=\dfrac{1-\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}\\&=\dfrac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1\end{align*}$
In the last step, divide each term on top by ${\sqrt{2}}$: ${\dfrac{2}{\sqrt{2}}=\sqrt{2}}$ and ${\dfrac{\sqrt{2}}{\sqrt{2}}=1}$.
Example 7: Suppose ${\cos u=-\dfrac{4}{5}}$ and ${\pi<u<\dfrac{3\pi}{2}}$. Find ${\sin\dfrac{u}{2}}$ and ${\cos\dfrac{u}{2}}$.
Solution:
Quadrant of ${\dfrac{u}{2}}$: Divide the inequality by ${2}$:
$\dfrac{\pi}{2}<\dfrac{u}{2}<\dfrac{3\pi}{4}$
So ${\dfrac{u}{2}}$ is in quadrant II. There sine is positive and cosine is negative.
$\begin{align*}\sin\dfrac{u}{2}&=+\sqrt{\dfrac{1-(-4/5)}{2}}=\sqrt{\dfrac{9}{10}}\\&=\dfrac{3}{\sqrt{10}}=\dfrac{3\sqrt{10}}{10}\end{align*}$
$\begin{align*}\cos\dfrac{u}{2}&=-\sqrt{\dfrac{1+(-4/5)}{2}}=-\sqrt{\dfrac{1}{10}}\\&=-\dfrac{1}{\sqrt{10}}=-\dfrac{\sqrt{10}}{10}\end{align*}$
Summary
- Double angle: ${\sin 2x=2\sin x\cos x}$ and ${\cos 2x=\cos^2 x-\sin^2 x}$, which also equals ${2\cos^2 x-1}$ and ${1-2\sin^2 x}$.
- Power reducing: ${\sin^2 x=\dfrac{1-\cos 2x}{2}}$ and ${\cos^2 x=\dfrac{1+\cos 2x}{2}}$.
- Half angle: ${\sin\dfrac{u}{2}=\pm\sqrt{\dfrac{1-\cos u}{2}}}$, with the sign from the quadrant of ${\dfrac{u}{2}}$.
← Sum and difference formulasProduct-to-sum and sum-to-product formulas →