Find the exact value of $\sin \frac{\pi}{12}$.

$\sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$

Find the exact value of $\cos \frac{7\pi}{12}$.

$\cos \frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$

Write $\sin 3x \cos x - \cos 3x \sin x$ as a single trig ratio, and evaluate it at $x = \frac{\pi}{8}$.

$\sin 2x$, which is $\frac{\sqrt{2}}{2}$ at $x = \frac{\pi}{8}$

Find the exact value of $\tan \frac{\pi}{12}$.

$\tan \frac{\pi}{12} = 2 - \sqrt{3}$

$\sin A = \frac{3}{5}$ with $\frac{\pi}{2} < A < \pi$, and $\cos B = \frac{5}{13}$ with $\frac{3\pi}{2} < B < 2\pi$. Find the exact values of $\sin(A + B)$ and $\cos(A - B)$.

$\sin(A + B) = \frac{63}{65}$ and $\cos(A - B) = -\frac{56}{65}$

Grade 12 · MHF4U · Advanced Functions

MHF4U: Compound Angle Formulas

$\sin(a + b)$ is not $\sin a + \sin b$ ; try $a = b = \frac{\pi}{2}$ and the left side is $0$ while the right side is $2$ . The compound angle formulas are what the sum really expands to, and they turn angles you do not know into angles you do.

How It Works

The six formulas, which you should be able to write from memory:

$\sin(a + b) = \sin a \cos b + \cos a \sin b$ and $\sin(a - b) = \sin a \cos b - \cos a \sin b$ .

$\cos(a + b) = \cos a \cos b - \sin a \sin b$ and $\cos(a - b) = \cos a \cos b + \sin a \sin b$ .

$\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$ and $\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$ .

Two things to notice. The sine formula keeps the sign of the angle and mixes sine with cosine. The cosine formula flips the sign and keeps like with like: cosines together, sines together. Learn those two patterns and the tangent formula is the only one left to memorise.

What they are for. First, exact values of angles that are sums or differences of special angles: $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ , $\frac{5\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4}$ , $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$ . Second, collapsing an expression that matches the right side of a formula back into a single ratio. Third, finding $\sin(a + b)$ from the ratios of $a$ and $b$ separately, where the work is in getting the missing ratios with the right signs from the quadrants.

The cofunction identities are a special case: $\cos\left(\frac{\pi}{2} - x\right) = \cos \frac{\pi}{2} \cos x + \sin \frac{\pi}{2} \sin x = \sin x$ . Any identity involving $\frac{\pi}{2} \pm x$ or $\pi \pm x$ can be proved the same way, by expanding and using the exact values of the special angle.

Always finish with a sign check. $\frac{7\pi}{12}$ is in quadrant 2, so its cosine must come out negative; if the surds in your answer give a positive number, a sign slipped in the expansion.

5 Worked Examples

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

Easy
Find the exact value of $\sin \frac{\pi}{12}$ .
Show solution
1. Write $\frac{\pi}{12}$ as a difference of special angles: $\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$ .
2. $\sin\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \sin \frac{\pi}{3} \cos \frac{\pi}{4} - \cos \frac{\pi}{3} \sin \frac{\pi}{4}$ .
3. $= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$ .
Answer: $\sin \frac{\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$
Easy
Find the exact value of $\cos \frac{7\pi}{12}$ .
Show solution
1. $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$ , since $\frac{4\pi}{12} + \frac{3\pi}{12} = \frac{7\pi}{12}$ .
2. $\cos\left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \cos \frac{\pi}{3} \cos \frac{\pi}{4} - \sin \frac{\pi}{3} \sin \frac{\pi}{4}$ .
3. $= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}$ .
4. Sign check: $\frac{7\pi}{12}$ is in quadrant 2, where cosine is negative, and $\sqrt{2} < \sqrt{6}$ makes the answer negative.
Answer: $\cos \frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$
Medium
Write $\sin 3x \cos x - \cos 3x \sin x$ as a single trig ratio, and evaluate it at $x = \frac{\pi}{8}$ .
Show solution
1. The expression has the shape $\sin a \cos b - \cos a \sin b$ with $a = 3x$ and $b = x$ , which is $\sin(a - b)$ .
2. So it equals $\sin(3x - x) = \sin 2x$ .
3. At $x = \frac{\pi}{8}$ : $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ .
Answer: $\sin 2x$ , which is $\frac{\sqrt{2}}{2}$ at $x = \frac{\pi}{8}$
Medium
Find the exact value of $\tan \frac{\pi}{12}$ .
Show solution
1. $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ , with $\tan \frac{\pi}{3} = \sqrt{3}$ and $\tan \frac{\pi}{4} = 1$ .
2. $\tan\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \cdot 1} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .
3. Rationalise by multiplying top and bottom by $\sqrt{3} - 1$ : $\frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2}$ .
Answer: $\tan \frac{\pi}{12} = 2 - \sqrt{3}$
Test level
$\sin A = \frac{3}{5}$ with $\frac{\pi}{2} < A < \pi$ , and $\cos B = \frac{5}{13}$ with $\frac{3\pi}{2} < B < 2\pi$ . Find the exact values of $\sin(A + B)$ and $\cos(A - B)$ .
Show solution
1. Find the missing ratios first. $A$ is in quadrant 2, so $\cos A < 0$ : $\cos A = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$ .
2. $B$ is in quadrant 4, so $\sin B < 0$ : $\sin B = -\sqrt{1 - \frac{25}{169}} = -\frac{12}{13}$ .
3. $\sin(A + B) = \sin A \cos B + \cos A \sin B = \frac{3}{5} \cdot \frac{5}{13} + \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) = \frac{15}{65} + \frac{48}{65} = \frac{63}{65}$ .
4. $\cos(A - B) = \cos A \cos B + \sin A \sin B = \left(-\frac{4}{5}\right)\frac{5}{13} + \frac{3}{5}\left(-\frac{12}{13}\right) = -\frac{20}{65} - \frac{36}{65} = -\frac{56}{65}$ .
Answer: $\sin(A + B) = \frac{63}{65}$ and $\cos(A - B) = -\frac{56}{65}$

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

Keep Going

Free Trigonometric Ratios practice test, with solutions

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