Lesson 2 of 5Trigonometric Identities & Equations

Unit 5 · Lesson 2 of 5

Compound Angle Formulas

Formulas for $sin (A \pm B)$ , $cos (A \pm B)$ and $tan (A \pm B)$ let you split unfamiliar angles like $\frac{π}{12}$ into familiar ones — and collapse long expressions back into a single ratio.

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Key things to remember

1 $sin (A \pm B) = sin A cos B \pm cos A sin B$ — mixed products, sign preserved.
2 $cos (A \pm B) = cos A cos B \mp sin A sin B$ — like products, sign FLIPPED.
3 $tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B} .$
4 $sin \frac{π}{12} = \frac{6 - 2}{4}$ and $cos \frac{π}{12} = \frac{6 + 2}{4}$ .
5 $sin \frac{5 π}{12} = \frac{6 + 2}{4}$ and $cos \frac{5 π}{12} = \frac{6 - 2}{4}$ .
6Read every formula backwards too: $sin X cos Y + cos X sin Y$ collapses to $sin (X + Y)$ .

The unit circle gives you exact values at a small set of angles: multiples of $\frac{π}{6}$ and $\frac{π}{4}$ . Compound angle formulas massively expand that list, because any angle you can write as a SUM or DIFFERENCE of two known angles now has an exact value too. They also run backwards, letting you compress a four-term expression into one trig ratio — which is often the whole point of a question.

The most common wrong assumption in the course

Sine is not a multiplier. You cannot distribute it over a sum. In general the sine of a sum is NOT the sum of the sines, and the same goes for cosine and tangent. Test it: the sine of ninety degrees is 1, but the sines of thirty and sixty degrees add to about 1.37.

The formulas

sin (A + B) = sin A cos B + cos A sin B sin (A - B) = sin A cos B - cos A sin B

Sine keeps the sign you were given: plus stays plus, minus stays minus.

cos (A + B) = cos A cos B - sin A sin B cos (A - B) = cos A cos B + sin A sin B

Cosine FLIPS the sign: plus becomes minus, minus becomes plus.

Read those two boxes side by side, because the difference between them is where almost all the marks are lost. The sine formulas MIX the ratios — every term is one sine times one cosine — and they keep the sign you were handed. The cosine formulas KEEP THE RATIOS TOGETHER — cosine with cosine, sine with sine — and they flip the sign. If you remember nothing else, remember that cosine flips.

Fig 1How each compound angle formula is assembled, and where the sign flip happens on the cosine line.

tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B} tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

Tangent: the numerator keeps the given sign, the denominator flips it.

VideoAn Ontario teacher proves the compound angle formulas, so you can see where the signs in each one come from rather than only memorizing them.Watch on YouTube

Exact values for new angles

To find an exact value for an unfamiliar angle, write it as a sum or difference of two angles you already know exactly. Work in twelfths of $π$ : since $\frac{π}{6} = \frac{2 π}{12}$ , $\frac{π}{4} = \frac{3 π}{12}$ and $\frac{π}{3} = \frac{4 π}{12}$ , combinations of these land on twelfths, which is why $\frac{π}{12}$ and $\frac{5 π}{12}$ are the classic exam angles.

Target angle	Decomposition	Check in twelfths
$\frac{π}{12}$	$\frac{π}{3} - \frac{π}{4}$	$\frac{4 π}{12} - \frac{3 π}{12} = \frac{π}{12}$
$\frac{5 π}{12}$	$\frac{π}{6} + \frac{π}{4}$	$\frac{2 π}{12} + \frac{3 π}{12} = \frac{5 π}{12}$
$\frac{7 π}{12}$	$\frac{π}{3} + \frac{π}{4}$	$\frac{4 π}{12} + \frac{3 π}{12} = \frac{7 π}{12}$
$\frac{11 π}{12}$	$\frac{2 π}{3} + \frac{π}{4}$	$\frac{8 π}{12} + \frac{3 π}{12} = \frac{11 π}{12}$

Standard decompositions. Convert to twelfths to check any decomposition you invent.

Fig 2Every exact value you feed into a compound angle formula comes from one of these two triangles.

VideoShort worked examples from the same Ontario teacher that apply the compound angle formulas.Watch on YouTube

Running the formulas backwards

The formulas are identities, so they are just as true read right to left. If you see an expression of the form $sin X cos Y + cos X sin Y$ , that IS $sin (X + Y)$ — collapse it. The tell is the pattern: two terms, each a sine times a cosine, with the two angles swapping roles between the terms. If instead both terms are cosine-cosine and sine-sine, you are looking at a cosine formula, and the sign between them tells you whether the angles are added or subtracted.

sin 5 x cos 3 x - cos 5 x sin 3 x = sin (5 x - 3 x) = sin 2 x

A typical simplify question: recognize the pattern, then collapse.

Sanity-check every exact value

After you produce an exact answer, punch both the original angle and your radical expression into a calculator in radian mode. They should agree to several decimal places. This catches a dropped sign in about five seconds.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Find the exact value of

sin \frac{π}{12}

Example 2

Find the exact value of

cos \frac{5 π}{12}

Example 3

Given

sin A = \frac{5}{13}

with

A

in quadrant I, and

cos B = - \frac{4}{5}

with

B

in quadrant II, find the exact value of

cos (A - B)

Example 4

Express

cos 7 x cos 2 x + sin 7 x sin 2 x

as a single trigonometric ratio.

Where marks get lost

Distributing: writing $cos (A + B) = cos A + cos B$ . There is no distributive law for trig functions — $cos$ is a function name, not a factor.
Getting the cosine sign backwards. $cos (A + B)$ has a MINUS between the products and $cos (A - B)$ has a PLUS. Sine is the one that keeps its sign.
Mixing the ratios in the cosine formula — writing $cos A sin B - sin A cos B$ . The cosine formula pairs like with like: cosine with cosine, sine with sine.
Decomposing an angle incorrectly, e.g. claiming $\frac{π}{12} = \frac{π}{6} - \frac{π}{4}$ . Convert to twelfths and check: that difference is $- \frac{π}{12}$ , not $\frac{π}{12}$ .
When given $sin A$ and asked for $cos (A + B)$ , forgetting that you must first find $cos A$ and $sin B$ — and that the quadrant, not the algebra, decides their signs.

Flashcards 10

sin (A + B)

sin (A - B)

cos (A + B)

cos (A - B)

+ 6 more

Practice

Which expansion of

cos (A + B)

is correct?

Which expansion of

sin (A - B)

is correct?

Which decomposition correctly produces

\frac{π}{12}

sin \frac{π}{12} = \frac{6 - 2}{4}

. Evaluate this to four decimal places.

Given

sin A = \frac{3}{5}

and

cos B = \frac{5}{13}

, with both

A

and

B

in quadrant I, find

sin (A + B)

as a decimal to four places.

tan \frac{5 π}{12} = 2 + 3

. Give its value to three decimal places.

Write

sin 5 x cos 3 x - cos 5 x sin 3 x

as a single trigonometric ratio.

True or false:

cos (A + B) = cos A + cos B

for all

A

and

B

Put these steps in order for finding the exact value of

cos \frac{7 π}{12}

1Apply $cos (A + B) = cos A cos B - sin A sin B$
2Combine to get $\frac{2 - 6}{4}$
3Substitute $cos \frac{π}{3} = \frac{1}{2}$ , $cos \frac{π}{4} = \frac{2}{2}$ , $sin \frac{π}{3} = \frac{3}{2}$ , $sin \frac{π}{4} = \frac{2}{2}$
4Write $\frac{7 π}{12} = \frac{π}{3} + \frac{π}{4}$

Match each expression to the single trigonometric ratio it collapses to.

A. $cos 3 x$
B. $cos 5 x$
C. $sin 3 x$
D. $sin 5 x$

sin 4 x cos x + cos 4 x sin x

cos 4 x cos x - sin 4 x sin x

cos 4 x cos x + sin 4 x sin x

sin 4 x cos x - cos 4 x sin x

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