Double-Angle and Half-Angle Formulas

Q: What are the double-angle formulas?

sin 2θ = 2 sin θ cos θ, cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ, and tan 2θ = 2 tan θ/(1 − tan²θ).

Q: Why isn't sin 2θ equal to 2 sin θ?

Sine is not linear. At θ = π/2, sin 2θ = sin π = 0 while 2 sin θ = 2. The correct formula carries a factor of cos θ.

Q: What are the half-angle formulas?

sin(θ/2) = ±√((1 − cos θ)/2) and cos(θ/2) = ±√((1 + cos θ)/2). The sign comes from the quadrant that θ/2 lies in.

Q: How do I pick the sign in a half-angle formula?

Work out which quadrant θ/2 is in, then take the sign that sine or cosine has there. The formula does not choose for you.

Q: What are the power-reducing formulas?

cos²θ = (1 + cos 2θ)/2 and sin²θ = (1 − cos 2θ)/2. They trade a square for a first power of a doubled angle.

Quick answer

Setting β = α in the sum formulas gives the double-angle formulas: sin 2θ = 2 sin θ cos θ and cos 2θ = cos²θ − sin²θ, which the Pythagorean identity rewrites as 2cos²θ − 1 or 1 − 2sin²θ. Solving those two versions for cos²θ and sin²θ gives the power-reducing formulas, and replacing θ by half an angle gives the half-angle formulas, whose sign comes from the quadrant of the half angle. Together they produce exact values such as cos 15° and turn equations with 2x in them into equations with x.

What you'll learn

Derive the double-angle formulas from the sum formulas
Choose the useful form of cos 2θ for a given problem
Find exact values with the half-angle formulas, including the sign
Solve equations that mix an angle with its double

Doubling an angle

The sum formulas already cover a double angle: $2\theta$ is $\theta + \theta$ . Setting $\beta = \alpha = \theta$ ,

\sin 2\theta = 2\sin\theta\cos\theta \qquad \cos 2\theta = \cos^2\theta - \sin^2\theta \qquad \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

The cosine formula has two more forms. Replacing $\sin^2\theta$ by $1 - \cos^2\theta$ , or $\cos^2\theta$ by $1 - \sin^2\theta$ , gives

\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta

Pick whichever form matches what you know. A problem that gives $\sin\theta$ wants the third version, since it never mentions the cosine.

y = sin 2x
y = 2 sin x

y = sin 2x and y = 2 sin x are different curves

Why halving reverses the same two formulas

Solve $\cos 2\theta = 1 - 2\sin^2\theta$ for $\sin^2\theta$ , and $\cos 2\theta = 2\cos^2\theta - 1$ for $\cos^2\theta$ :

\sin^2\theta = \frac{1 - \cos 2\theta}{2} \qquad \cos^2\theta = \frac{1 + \cos 2\theta}{2}

These are the power-reducing formulas: a square on the left, a first power on the right. Now write $\theta = \tfrac{\alpha}{2}$ , so $2\theta = \alpha$ , and take square roots:

\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}} \qquad \cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}

The double-angle and half-angle formulas are the same two statements read in opposite directions, which is why the half-angle versions carry a square root and a sign. A square root is never negative, so the sign has to be supplied: decide which quadrant $\tfrac{\alpha}{2}$ falls in and take the sign that sine or cosine has there.

Worked examples

Example 3 — Choosing the sign

Given $\cos\alpha = -\tfrac{3}{5}$ with $\alpha$ in the second quadrant, find $\sin\tfrac{\alpha}{2}$ and $\cos\tfrac{\alpha}{2}$ .

In the second quadrant $90° < \alpha < 180°$ , so $45° < \tfrac{\alpha}{2} < 90°$ : the half angle is in the first quadrant, where both are positive.

\sin\frac{\alpha}{2} = \sqrt{\frac{1 + \tfrac{3}{5}}{2}} = \sqrt{\tfrac{4}{5}} = \frac{2\sqrt{5}}{5} \qquad \cos\frac{\alpha}{2} = \sqrt{\frac{1 - \tfrac{3}{5}}{2}} = \sqrt{\tfrac{1}{5}} = \frac{\sqrt{5}}{5}

Their quotient gives $\tan\tfrac{\alpha}{2} = 2$ .

Common mistakes

Practice problems

Given $\sin\theta = \tfrac{4}{5}$ with $\theta$ in the first quadrant, find $\sin 2\theta$ and $\cos 2\theta$ .

Answer

$\sin 2\theta = \tfrac{24}{25}$ and $\cos 2\theta = -\tfrac{7}{25}$

Full solution

$\cos\theta = \tfrac{3}{5}$ , so $\sin 2\theta = 2 \cdot \tfrac{4}{5} \cdot \tfrac{3}{5} = \tfrac{24}{25}$ and $\cos 2\theta = 1 - 2\left(\tfrac{16}{25}\right) = -\tfrac{7}{25}$ .
Given $\cos\theta = \tfrac{5}{13}$ with $\theta$ in the fourth quadrant, find $\sin 2\theta$ and $\cos 2\theta$ .

Answer

$\sin 2\theta = -\tfrac{120}{169}$ and $\cos 2\theta = -\tfrac{119}{169}$

Full solution

In the fourth quadrant $\sin\theta = -\tfrac{12}{13}$ . Then $\sin 2\theta = 2\left(-\tfrac{12}{13}\right)\left(\tfrac{5}{13}\right)$ and $\cos 2\theta = 2\left(\tfrac{25}{169}\right) - 1$ .
Find $\cos 22.5°$ exactly.

Answer

$\tfrac{\sqrt{2 + \sqrt{2}}}{2} \approx 0.924$

Full solution

Half of $45°$ , with $\cos 45° = \tfrac{\sqrt{2}}{2}$ . The angle is in the first quadrant, so the sign is positive: $\sqrt{\tfrac{1 + \sqrt{2}/2}{2}} = \tfrac{\sqrt{2 + \sqrt{2}}}{2}$ .
Find $\sin 75°$ exactly with a half-angle formula.

Answer

$\tfrac{\sqrt{2 + \sqrt{3}}}{2} \approx 0.966$

Full solution

$75°$ is half of $150°$ , and $\cos 150° = -\tfrac{\sqrt{3}}{2}$ . Sine is positive in the first quadrant: $\sqrt{\tfrac{1 + \sqrt{3}/2}{2}} = \tfrac{\sqrt{2 + \sqrt{3}}}{2}$ .
Solve $\cos 2x = \cos x$ for $0 \le x < 2\pi$ .

Answer

$x = 0$ , $x = \tfrac{2\pi}{3}$ , $x = \tfrac{4\pi}{3}$

Full solution

$2\cos^2 x - \cos x - 1 = 0$ factors as $(2\cos x + 1)(\cos x - 1) = 0$ . Then $\cos x = -\tfrac{1}{2}$ gives $\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$ , and $\cos x = 1$ gives $0$ .
Verify $\displaystyle\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$ .

Answer

Both sides equal $\cos^2 x - \sin^2 x$ .

Full solution

The denominator is $\sec^2 x$ , so the right side is $\left(1 - \tan^2 x\right)\cos^2 x = \cos^2 x - \sin^2 x$ , which is $\cos 2x$ .
Write $\sin 4x$ in terms of functions of $2x$ .

Answer

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

Full solution

Apply the double-angle formula to the angle $2x$ : $4x$ is its double.
Given $\cos\theta = \tfrac{7}{25}$ with $\theta$ in the first quadrant, find $\tan\tfrac{\theta}{2}$ .

Answer

$\tfrac{3}{4}$

Full solution

$\sin\theta = \tfrac{24}{25}$ , and $\tan\tfrac{\theta}{2} = \tfrac{1 - \cos\theta}{\sin\theta} = \tfrac{18/25}{24/25}$ .
Show that $\sin^2 x\cos^2 x = \tfrac{1 - \cos 4x}{8}$ .

Hint

Start with $(\sin x \cos x)^2$ and use a double angle twice.

Answer

$(\sin x\cos x)^2 = \tfrac{1}{4}\sin^2 2x = \tfrac{1}{4} \cdot \tfrac{1 - \cos 4x}{2}$

Full solution

Since $\sin 2x = 2\sin x\cos x$ , the product $\sin x \cos x$ is $\tfrac{1}{2}\sin 2x$ , so its square is $\tfrac{1}{4}\sin^2 2x$ . The power-reducing formula with the angle $2x$ gives $\sin^2 2x = \tfrac{1 - \cos 4x}{2}$ .
A student writes $\sin 2x = 2\sin x$ and solves $\sin 2x = 1$ as $\sin x = \tfrac{1}{2}$ . What went wrong?

Hint

Test the student’s answer in the original equation.

Answer

$\sin 2x = 2\sin x\cos x$ , not $2\sin x$ . The equation $\sin 2x = 1$ gives $x = \tfrac{\pi}{4}$ and $x = \tfrac{5\pi}{4}$ .

Full solution

The student’s answer $x = \tfrac{\pi}{6}$ gives $\sin\tfrac{\pi}{3} \approx 0.866$ , not $1$ . Solving properly: $2x = \tfrac{\pi}{2} + 2\pi k$ , so $x = \tfrac{\pi}{4} + \pi k$ , which lands on $\tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$ in one turn.

Frequently asked questions

What are the double-angle formulas?

sin 2θ = 2 sin θ cos θ, cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ, and tan 2θ = 2 tan θ/(1 − tan²θ).

Why isn't sin 2θ equal to 2 sin θ?