If $x = \frac{\pi}{6}$, find $\sin 2x$ and $\cos 2x$ using the double angle formulas, and confirm against the unit circle.

$\sin 2x = \frac{\sqrt{3}}{2}$ and $\cos 2x = \frac{1}{2}$

If $\cos x = \frac{3}{5}$ and $0 < x < \frac{\pi}{2}$, find the exact values of $\sin 2x$ and $\cos 2x$.

$\sin 2x = \frac{24}{25}$ and $\cos 2x = -\frac{7}{25}$

Write each as a single trig ratio: $6 \sin x \cos x$, $\cos^2 4x - \sin^2 4x$ and $1 - 2\sin^2 \frac{x}{2}$.

$3 \sin 2x$, $\cos 8x$ and $\cos x$

If $\tan x = -2$ and $\frac{\pi}{2} < x < \pi$, find the exact values of $\tan 2x$, $\sin 2x$ and $\cos 2x$.

$\tan 2x = \frac{4}{3}$, $\sin 2x = -\frac{4}{5}$, $\cos 2x = -\frac{3}{5}$

Solve $\sin 2x = \sin x$ for $0 \le x < 2\pi$.

$x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Double Angle Formulas

Put $b = a$ into a compound angle formula and you get a double angle formula: $\sin 2x$ in terms of $\sin x$ and $\cos x$ . They let you move between an angle and twice that angle, which is what most trig equations and identities with a $2x$ in them need.

How It Works

From $\sin(x + x) = \sin x \cos x + \cos x \sin x$ : $\sin 2x = 2 \sin x \cos x$ .

From $\cos(x + x) = \cos x \cos x - \sin x \sin x$ : $\cos 2x = \cos^2 x - \sin^2 x$ . Using $\sin^2 x + \cos^2 x = 1$ to replace one of the squares gives the other two forms, $\cos 2x = 2\cos^2 x - 1$ and $\cos 2x = 1 - 2\sin^2 x$ .

From the tangent sum formula: $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$ .

Which cosine form? The one that matches what you have. If the question gives $\cos x$ , use $2\cos^2 x - 1$ and you never need $\sin x$ . If it gives $\sin x$ , use $1 - 2\sin^2 x$ . If the equation you are solving has a $\cos x$ elsewhere in it, pick the form that keeps everything in cosine, so that the result factors.

Reading them backwards matters as much as forwards. $6 \sin x \cos x$ is $3 \sin 2x$ ; $1 - 2\sin^2 4x$ is $\cos 8x$ ; $\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ is $\cos x$ . The angle in the formula is whatever is being doubled, not always a plain $x$ .

Given one ratio, find the doubles. Draw the angle in its quadrant, find the missing ratio with the right sign, then substitute. The sign of $\sin 2x$ and $\cos 2x$ comes out of the formula; do not guess it from the quadrant of $x$ , because $2x$ is usually in a different quadrant.

In an equation like $\sin 2x = \sin x$ , expand the double angle, move everything to one side and factor. Dividing both sides by $\sin x$ instead throws away the solutions where $\sin x = 0$ .

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
If $x = \frac{\pi}{6}$ , find $\sin 2x$ and $\cos 2x$ using the double angle formulas, and confirm against the unit circle.
Show solution
1. $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ .
2. $\sin 2x = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$ .
3. $\cos 2x = 2\cos^2 x - 1 = 2 \cdot \frac{3}{4} - 1 = \frac{1}{2}$ .
4. $2x = \frac{\pi}{3}$ , and the unit circle gives $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$ . They match.
Answer: $\sin 2x = \frac{\sqrt{3}}{2}$ and $\cos 2x = \frac{1}{2}$
Easy
If $\cos x = \frac{3}{5}$ and $0 < x < \frac{\pi}{2}$ , find the exact values of $\sin 2x$ and $\cos 2x$ .
Show solution
1. Quadrant 1, so $\sin x > 0$ : $\sin x = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ . A $3, 4, 5$ triangle.
2. $\sin 2x = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$ .
3. Cosine is given, so use $\cos 2x = 2\cos^2 x - 1 = 2 \cdot \frac{9}{25} - 1 = -\frac{7}{25}$ .
4. The negative cosine says $2x$ is past $\frac{\pi}{2}$ , which fits: $x$ is a little under $1$ radian, so $2x$ is in quadrant 2.
Answer: $\sin 2x = \frac{24}{25}$ and $\cos 2x = -\frac{7}{25}$
Medium
Write each as a single trig ratio: $6 \sin x \cos x$ , $\cos^2 4x - \sin^2 4x$ and $1 - 2\sin^2 \frac{x}{2}$ .
Show solution
1. $6 \sin x \cos x = 3(2 \sin x \cos x) = 3 \sin 2x$ .
2. $\cos^2 4x - \sin^2 4x$ is $\cos 2\theta$ with $\theta = 4x$ , so it is $\cos 8x$ .
3. $1 - 2\sin^2 \frac{x}{2}$ is $\cos 2\theta$ with $\theta = \frac{x}{2}$ , so it is $\cos x$ .
Answer: $3 \sin 2x$ , $\cos 8x$ and $\cos x$
Medium
If $\tan x = -2$ and $\frac{\pi}{2} < x < \pi$ , find the exact values of $\tan 2x$ , $\sin 2x$ and $\cos 2x$ .
Show solution
1. $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x} = \frac{-4}{1 - 4} = \frac{4}{3}$ .
2. For the others, draw $x$ in quadrant 2: opposite $2$ , adjacent $-1$ , hypotenuse $\sqrt{5}$ . So $\sin x = \frac{2}{\sqrt{5}}$ and $\cos x = -\frac{1}{\sqrt{5}}$ .
3. $\sin 2x = 2 \cdot \frac{2}{\sqrt{5}} \cdot \left(-\frac{1}{\sqrt{5}}\right) = -\frac{4}{5}$ .
4. $\cos 2x = \cos^2 x - \sin^2 x = \frac{1}{5} - \frac{4}{5} = -\frac{3}{5}$ .
5. Check: $\frac{\sin 2x}{\cos 2x} = \frac{-4/5}{-3/5} = \frac{4}{3} = \tan 2x$ .
Answer: $\tan 2x = \frac{4}{3}$ , $\sin 2x = -\frac{4}{5}$ , $\cos 2x = -\frac{3}{5}$
Test level
Solve $\sin 2x = \sin x$ for $0 \le x < 2\pi$ .
Show solution
1. Expand the double angle: $2 \sin x \cos x = \sin x$ .
2. Move everything to one side and factor; do not divide by $\sin x$ : $2 \sin x \cos x - \sin x = 0$ , so $\sin x (2 \cos x - 1) = 0$ .
3. $\sin x = 0$ at $x = 0$ and $x = \pi$ .
4. $\cos x = \frac{1}{2}$ at $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$ .
5. Dividing by $\sin x$ would have left only $\cos x = \frac{1}{2}$ and lost two of the four solutions.
Answer: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$

Learn It on Martian Lab

Martian Lab teaches this in the Double Compound Angle Formula lesson of its MHF4U Trigonometric Ratios unit. Watch it, then take the Trigonometric Ratios practice tests to see if it stuck.

Watch the Double Compound Angle Formula lesson free

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Free Trigonometric Ratios practice test, with solutions

More Trigonometric Ratios topics: Proving Trig Identities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas

More MHF4U topics: Long and Synthetic Division · Vertical and Horizontal Asymptotes · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Remainder and Factor Theorem · Solving Polynomial Equations · Polynomial Inequalities · Sum and Difference of Cubes · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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