Compound-angle identities | Year 11

01 / LEARN

Understand the idea.

Combine angles using products

Trigonometric functions do not distribute over addition. Use the compound-angle identities to replace a sine or cosine of a sum with products of known ratios. For cosine, a sum uses subtraction between the products, while a difference uses addition.

Choose component angles with exact known values, such as $75^{\circ} = 45^{\circ} + 30^{\circ}$ . Write the identity before substituting and keep fractions and surds exact. Combine terms only after the correct products and signs are visible.

sin (A \pm B) = sin A cos B \pm cos A sin B

Recover missing information first

If only one ratio is given, recover the other from ${sin}^{2} A + {cos}^{2} A = 1$ . Taking a square root gives a magnitude; the stated quadrant determines its sign. Without enough quadrant information, more than one result may be possible.

The tangent sum formula comes from dividing the sine sum formula by the cosine sum formula. Use it only where the component tangents and resulting denominator are defined. A zero denominator means an undefined tangent, not a finite value of large magnitude.

cos (A \pm B) = cos A cos B \mp sin A sin B

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Example 1

Calculate $sin 75^{\circ}$ exactly.

Worked solution

1
Split the angle and choose the identity.
$sin 75^{\circ} = sin 45^{\circ} cos 30^{\circ} + cos 45^{\circ} sin 30^{\circ}$
The sine of a sum is the sum of the two cross-products.
2
Insert the exact special-angle values.
$sin 75^{\circ} = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2}$
The component angles are acute, so all four ratios are positive.
3
Multiply and combine over a common denominator.
$sin 75^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$
Both products have denominator four and their exact numerators add.

Answer

$(\sqrt{6} + \sqrt{2}) / 4$ .

EXAMPLE 02

Example 2

For acute $A, B$ , $sin A = 3 / 5$ and $cos B = 12 / 13$ . Find $cos (A + B)$ .

Worked solution

1
Recover the missing positive ratios.
$\begin{aligned} cos A & = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \\ sin B & = \sqrt{1 - \frac{144}{169}} = \frac{5}{13} \end{aligned}$
Both angles are acute, so the positive square roots are required.
2
Apply the cosine sum identity.
$cos (A + B) = cos A cos B - sin A sin B$
Cosine of a sum subtracts the sine product from the cosine product.
3
Substitute and simplify.
$\begin{aligned} cos (A + B) & = \frac{4}{5} \frac{12}{13} - \frac{3}{5} \frac{5}{13} \\ = \frac{48 - 15}{65} = \frac{33}{65} \end{aligned}$
The products share denominator sixty-five, so their numerators subtract.

Answer

$33 / 65$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Calculate $cos (45^{\circ} - 30^{\circ})$ exactly.

Show answer & reasoning

$(\sqrt{6} + \sqrt{2}) / 4$ .

Use $cos (A - B) = cos A cos B + sin A sin B$ . Substitution gives $\sqrt{6} / 4 + \sqrt{2} / 4 = (\sqrt{6} + \sqrt{2}) / 4$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find $sin 15^{\circ}$ exactly.
2 marks
Find $cos 75^{\circ}$ exactly.
2 marks
Find $tan 75^{\circ}$ exactly.
2 marks
For acute $A, B$ with $sin A = \frac{5}{13}$ and $sin B = \frac{3}{5}$ , find $sin (A + B)$ .
2 marks
Given $tan A = 2$ , $tan B = \frac{1}{3}$ , calculate $tan (A - B)$ .
2 marks
Disprove $cos (A + B) = cos A + cos B$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $sin 15^{\circ}$ exactly.
2 marks
Expand $sin (45^{\circ} - 30^{\circ})$ .
$\frac{\sqrt{6} - \sqrt{2}}{4}$ .
Find $cos 75^{\circ}$ exactly.
2 marks
Expand $cos (45^{\circ} + 30^{\circ})$ .
$\frac{\sqrt{6} - \sqrt{2}}{4}$ .
Find $tan 75^{\circ}$ exactly.
2 marks
Use $\frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$ , then rationalise.
$2 + \sqrt{3}$ .
For acute $A, B$ with $sin A = \frac{5}{13}$ and $sin B = \frac{3}{5}$ , find $sin (A + B)$ .
2 marks
The missing cosines are $\frac{12}{13}$ and $\frac{4}{5}$ .
Add $\frac{20}{65} + \frac{36}{65}$ .
$\frac{56}{65}$ .
Given $tan A = 2$ , $tan B = \frac{1}{3}$ , calculate $tan (A - B)$ .
2 marks
$\frac{2 - \frac{1}{3}}{1 + \frac{2}{3}} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1$ .
1.
Disprove $cos (A + B) = cos A + cos B$ .
2 marks
Both sides are defined, and the unequal values disprove the universal claim.
Set $A = B = 0$ : the two sides are 1 and 2.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Further Trigonometry: periodic functions and models, trigonometric identities and reciprocal functions; the Library sequence follows the approved school-topic structure.

Official curriculum source ↗

In this lesson

Understand the idea.

Combine angles using products

Recover missing information first

See the method in action.

Example 1

Split the angle and choose the identity.

Insert the exact special-angle values.

Multiply and combine over a common denominator.

Example 2

Recover the missing positive ratios.

Apply the cosine sum identity.

Substitute and simplify.

Try it yourself.

Check your understanding

Keep practising