Double Angle Formulas

What happens to the sine of an angle when you double the angle? It does not double. The double angle formulas tell you exactly what happens. They’re a special case of the compound angle formulas, and they show up constantly: in proving identities, in solving equations, and later in calculus and physics. All angles are in radians.

Key ideas

Deriving the formulas

Put $b = a$ in the compound angle formulas, and call the angle $x$.

Sine:

$$\sin 2x = \sin(x + x) = \sin x\cos x + \cos x\sin x = 2\sin x\cos x$$

Cosine:

$$\cos 2x = \cos(x + x) = \cos x\cos x - \sin x\sin x = \cos^2 x - \sin^2 x$$

Tangent:

$$\tan 2x = \tan(x + x) = \frac{\tan x + \tan x}{1 - \tan x\tan x} = \frac{2\tan x}{1 - \tan^2 x}$$

Three forms of cos 2x

Use $\sin^2 x + \cos^2 x = 1$ to replace one of the squares in $\cos^2 x - \sin^2 x$:

$$\begin{aligned} \cos 2x &= \cos^2 x - \sin^2 x \\ &= \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1 \\ &= (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x \end{aligned}$$

Choose the form that suits the problem: $2\cos^2 x - 1$ when you only want cosines, $1 - 2\sin^2 x$ when you only want sines.

Summary
$\sin 2x = 2\sin x\cos x$	$\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}$
$\cos 2x = \cos^2 x - \sin^2 x$	$\cos 2x = 2\cos^2 x - 1$
$\cos 2x = 1 - 2\sin^2 x$

Read the notation carefully

$\sin 2x$ means $\sin(2x)$: double the angle, then take the sine. It is not $2\sin x$. With $x = \dfrac{\pi}{2}$: $\sin 2x = \sin\pi = 0$, but $2\sin x = 2\sin\dfrac{\pi}{2} = 2$.

The formulas work for any angle that is “double” another one. For example, $\sin 6x = 2\sin 3x\cos 3x$ and $\cos x = 1 - 2\sin^2\dfrac{x}{2}$.

Finding sin 2x from one ratio and a quadrant

If you know one ratio of $x$ and its quadrant:

1. Sketch $x$ in standard position and draw a right triangle to the $x$-axis.
2. Use the Pythagorean theorem to find the missing side, with signs from the quadrant.
3. Read off $\sin x$, $\cos x$ and $\tan x$.
4. Substitute into the double angle formulas.

Worked examples

Example 1: Collapsing to a single ratio

Write each expression as a single trig ratio, then find its exact value.

(a) $2\sin\dfrac{\pi}{8}\cos\dfrac{\pi}{8}$

(b) $\cos^2\dfrac{\pi}{12} - \sin^2\dfrac{\pi}{12}$

(c) $\dfrac{2\tan\frac{\pi}{6}}{1 - \tan^2\frac{\pi}{6}}$

Solution.

(a) This is $\sin 2x$ with $x = \dfrac{\pi}{8}$: $\quad\sin\left(2\cdot\dfrac{\pi}{8}\right) = \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$

(b) This is $\cos 2x$ with $x = \dfrac{\pi}{12}$: $\quad\cos\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}$

(c) This is $\tan 2x$ with $x = \dfrac{\pi}{6}$: $\quad\tan\dfrac{\pi}{3} = \sqrt{3}$

Check (c) directly: $\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$, so the expression is $\dfrac{2/\sqrt{3}}{1 - 1/3} = \dfrac{2}{\sqrt{3}}\cdot\dfrac{3}{2} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$. ✓

Example 2: Simplifying

Simplify each expression.

(a) $1 - 2\sin^2 3x$

(b) $4\sin x\cos x$

Solution.

(a) This is $1 - 2\sin^2\theta$ with $\theta = 3x$, which equals $\cos 2\theta$:

$$1 - 2\sin^2 3x = \cos 6x$$

(b) Split off the $2$ that belongs to the formula:

$$4\sin x\cos x = 2(2\sin x\cos x) = 2\sin 2x$$

Example 3: Using a quadrant

Given $\cos x = -\dfrac{3}{5}$ and $\dfrac{\pi}{2} \lt x \lt \pi$, find the exact values of $\sin 2x$, $\cos 2x$ and $\tan 2x$. In which quadrant is $2x$?

Solution. $x$ is in quadrant 2. Draw a triangle with adjacent side $-3$ and hypotenuse $5$. The opposite side is $\sqrt{5^2 - 3^2} = 4$, and it’s positive because the point is above the $x$-axis.

For $\cos x = -\dfrac{3}{5}$ in quadrant 2, the terminal arm passes through $(-3, 4)$.

So $\sin x = \dfrac{4}{5}$, $\cos x = -\dfrac{3}{5}$, and $\tan x = -\dfrac{4}{3}$.

$$\begin{aligned} \sin 2x &= 2\sin x\cos x = 2\left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25} \\[4pt] \cos 2x &= 2\cos^2 x - 1 = 2\left(\frac{9}{25}\right) - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25} \\[4pt] \tan 2x &= \frac{\sin 2x}{\cos 2x} = \frac{-24/25}{-7/25} = \frac{24}{7} \end{aligned}$$

Check $\tan 2x$ with its own formula: $\dfrac{2\left(-\frac{4}{3}\right)}{1 - \frac{16}{9}} = \dfrac{-\frac{8}{3}}{-\frac{7}{9}} = \dfrac{8}{3}\cdot\dfrac{9}{7} = \dfrac{24}{7}$. ✓

Both $\sin 2x$ and $\cos 2x$ are negative, so $2x$ is in quadrant 3. That fits: $\pi \lt 2x \lt 2\pi$, and $\tan 2x$ is positive in quadrant 3.

Example 4: Choosing the right form of cos 2x

Prove that $\dfrac{\sin 2x}{1 + \cos 2x} = \tan x$.

Solution. In the denominator, the form $\cos 2x = 2\cos^2 x - 1$ cancels the $1$:

$$\begin{aligned} \text{L.S.} &= \frac{\sin 2x}{1 + \cos 2x} \\ &= \frac{2\sin x\cos x}{1 + 2\cos^2 x - 1} \\ &= \frac{2\sin x\cos x}{2\cos^2 x} \\ &= \frac{\sin x}{\cos x} \\ &= \tan x = \text{R.S.} \end{aligned}$$

A quick check (not a proof) with $x = \dfrac{\pi}{6}$: $\dfrac{\sin\frac{\pi}{3}}{1 + \cos\frac{\pi}{3}} = \dfrac{\sqrt{3}/2}{3/2} = \dfrac{\sqrt{3}}{3}$, and $\tan\dfrac{\pi}{6} = \dfrac{\sqrt{3}}{3}$. ✓

Common mistakes

Writing $\sin 2x = 2\sin x$. The $2$ is inside the function. Try $x = \dfrac{\pi}{2}$: $\sin\pi = 0$, but $2\sin\dfrac{\pi}{2} = 2$. The correct formula is $\sin 2x = 2\sin x\cos x$.

Mixing up the forms of cos 2x. It’s $1 - 2\sin^2 x$ and $2\cos^2 x - 1$, not $2\sin^2 x - 1$. A quick test: at $x = 0$, $\cos 0 = 1$, and $1 - 2\sin^2 0 = 1$, but $2\sin^2 0 - 1 = -1$.

Forgetting the quadrant when finding a missing ratio. In Example 3, $\sqrt{25 - 9}$ gives the size $4$, but the quadrant decides the sign. Draw the triangle in the correct quadrant first.

Using the quadrant of x for 2x. $x$ and $2x$ are usually in different quadrants. Get the signs of $\sin 2x$ and $\cos 2x$ from the formulas, not by guessing.

Losing the 2 when reversing a formula. $\sin x\cos x$ is $\dfrac{1}{2}\sin 2x$, not $\sin 2x$. Only $2\sin x\cos x$ collapses to $\sin 2x$.

Practice

1. (Warm-up) Write each as a single trig ratio. Where the angle is a special angle, give the exact value.

• (a) $2\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12}$
• (b) $\cos^2\dfrac{\pi}{8} - \sin^2\dfrac{\pi}{8}$
• (c) $2\cos^2 5x - 1$
• (d) $6\sin 3x\cos 3x$

Solution

(a) $\sin\dfrac{2\pi}{12} = \sin\dfrac{\pi}{6} = \dfrac{1}{2}$

(b) $\cos\dfrac{2\pi}{8} = \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$

(c) $\cos 10x$

(d) $3(2\sin 3x\cos 3x) = 3\sin 6x$

2. (Warm-up) Use $x = \dfrac{\pi}{4}$ to show that $\sin 2x = 2\sin x$ is not an identity.

Solution

Left side: $\sin\dfrac{\pi}{2} = 1$. Right side: $2\sin\dfrac{\pi}{4} = 2\cdot\dfrac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$.

Since $1 \ne \sqrt{2}$, the equation is false for $x = \dfrac{\pi}{4}$, so it isn’t an identity.

3. (Core) Given $\sin x = \dfrac{5}{13}$ and $0 \lt x \lt \dfrac{\pi}{2}$, find the exact values of $\sin 2x$, $\cos 2x$ and $\tan 2x$.

Solution

In quadrant 1, $\cos x = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}$.

$$\begin{aligned} \sin 2x &= 2\left(\frac{5}{13}\right)\left(\frac{12}{13}\right) = \frac{120}{169} \\[4pt] \cos 2x &= 1 - 2\left(\frac{25}{169}\right) = \frac{169 - 50}{169} = \frac{119}{169} \\[4pt] \tan 2x &= \frac{120/169}{119/169} = \frac{120}{119} \end{aligned}$$

4. (Core) Given $\sin x = -\dfrac{2}{3}$ and $\dfrac{3\pi}{2} \lt x \lt 2\pi$, find the exact values of $\sin 2x$ and $\cos 2x$.

Solution

In quadrant 4, cosine is positive: $\cos x = \sqrt{1 - \dfrac{4}{9}} = \sqrt{\dfrac{5}{9}} = \dfrac{\sqrt{5}}{3}$.

$$\sin 2x = 2\left(-\frac{2}{3}\right)\left(\frac{\sqrt{5}}{3}\right) = -\frac{4\sqrt{5}}{9}$$$$\cos 2x = 1 - 2\sin^2 x = 1 - 2\left(\frac{4}{9}\right) = \frac{1}{9}$$

5. (Core) Given $\tan x = 3$ and $\pi \lt x \lt \dfrac{3\pi}{2}$, find the exact values of $\tan 2x$, $\sin 2x$ and $\cos 2x$.

Solution

$$\tan 2x = \frac{2(3)}{1 - 3^2} = \frac{6}{-8} = -\frac{3}{4}$$

In quadrant 3 both sine and cosine are negative. With opposite $-3$, adjacent $-1$ and hypotenuse $\sqrt{9 + 1} = \sqrt{10}$: $\sin x = -\dfrac{3}{\sqrt{10}}$ and $\cos x = -\dfrac{1}{\sqrt{10}}$.

$$\sin 2x = 2\left(-\frac{3}{\sqrt{10}}\right)\left(-\frac{1}{\sqrt{10}}\right) = \frac{6}{10} = \frac{3}{5}$$$$\cos 2x = \cos^2 x - \sin^2 x = \frac{1}{10} - \frac{9}{10} = -\frac{4}{5}$$

Check: $\dfrac{\sin 2x}{\cos 2x} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$. ✓

6. (Core) Prove that $(\sin x + \cos x)^2 = 1 + \sin 2x$.

Solution

$$\begin{aligned} \text{L.S.} &= \sin^2 x + 2\sin x\cos x + \cos^2 x \\ &= (\sin^2 x + \cos^2 x) + 2\sin x\cos x \\ &= 1 + \sin 2x = \text{R.S.} \end{aligned}$$

7. (Core) Prove that $\cos^4 x - \sin^4 x = \cos 2x$.

Solution

Factor the left side as a difference of squares:

$$\begin{aligned} \text{L.S.} &= (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) \\ &= (\cos^2 x - \sin^2 x)(1) \\ &= \cos 2x = \text{R.S.} \end{aligned}$$

8. (Challenge) Use $\cos 2x = 2\cos^2 x - 1$ with $x = \dfrac{\pi}{8}$ to find the exact value of $\cos\dfrac{\pi}{8}$.

Solution

With $x = \dfrac{\pi}{8}$, $2x = \dfrac{\pi}{4}$:

$$\begin{aligned} \cos\frac{\pi}{4} &= 2\cos^2\frac{\pi}{8} - 1 \\ \frac{\sqrt{2}}{2} + 1 &= 2\cos^2\frac{\pi}{8} \\ \cos^2\frac{\pi}{8} &= \frac{2 + \sqrt{2}}{4} \end{aligned}$$

$\dfrac{\pi}{8}$ is in quadrant 1, so take the positive root:

$$\cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$$

Check: $\dfrac{\sqrt{2 + 1.4142}}{2} \approx 0.9239$, which matches $\cos\dfrac{\pi}{8}$ on a calculator. ✓

9. (Challenge) Write $\cos 3x$ in terms of $\cos x$ only. (Hint: $3x = 2x + x$.)

Solution

$$\begin{aligned} \cos 3x &= \cos(2x + x) \\ &= \cos 2x\cos x - \sin 2x\sin x \\ &= (2\cos^2 x - 1)\cos x - (2\sin x\cos x)\sin x \\ &= 2\cos^3 x - \cos x - 2\sin^2 x\cos x \\ &= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x \\ &= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x \\ &= 4\cos^3 x - 3\cos x \end{aligned}$$

Check with $x = 0$: $\cos 0 = 1$ and $4 - 3 = 1$. ✓ With $x = \dfrac{\pi}{3}$: $\cos\pi = -1$ and $4\left(\dfrac{1}{8}\right) - 3\left(\dfrac{1}{2}\right) = \dfrac{1}{2} - \dfrac{3}{2} = -1$. ✓