Double-angle identities and derived forms | Year 11

01 / LEARN

The idea, step by step

Set the angles equal

Substituting the same angle twice gives $sin 2 x = 2 sin x cos x$ and $cos 2 x = {cos}^{2} x - {sin}^{2} x$ . The Pythagorean identity produces two further cosine forms; choose the one using the ratio already known.

The tangent formula $tan 2 x = 2 tan x / (1 - {tan}^{2} x)$ requires the ratios to exist. When its denominator is zero, examine the original angle rather than trying to divide by zero.

cos 2 x = 1 - 2 {sin}^{2} x = 2 {cos}^{2} x - 1

Rearrange to handle squares

The cosine identities give ${sin}^{2} x = (1 - cos 2 x) / 2$ and ${cos}^{2} x = (1 + cos 2 x) / 2$ . These rewrite squares and will be useful in later integration. They do not by themselves give the sign of sine or cosine.

When taking square roots for a half-angle, determine its quadrant before choosing positive or negative. Retain exact square-root forms where a decimal is not required.

{sin}^{2} x = \frac{1 - cos 2 x}{2}, {cos}^{2} x = \frac{1 + cos 2 x}{2}

A common trap

Choosing a positive square root automatically.

Instead: The square gives a magnitude; the quadrant determines the sign.

02 / WORKED EXAMPLES

See the thinking

Example 1

For acute $x$ with $sin x = 5 / 13$ , find $sin 2 x$ and $cos 2 x$ .

The missing cosine is $12 / 13$ .
Double-angle sine is $2 (5 / 13) (12 / 13) = 120 / 169$ .
Double-angle cosine is $1 - 2 (25 / 169) = 119 / 169$ .

sin 2 x = 120 / 169

and

cos 2 x = 119 / 169

Example 2

Find $sin (π / 8)$ exactly.

Use the square identity: $(1 - cos (π / 4)) / 2 = (2 - \sqrt{2}) / 4$ .
Since $π / 8$ is acute, sine is positive.
Take the positive square root.

\sqrt{2 - \sqrt{2}} / 2

Check your understanding

Simplify $1 - 2 {sin}^{2} x$ .

Show answer

$cos 2 x$ .

This is one cosine double-angle identity.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

If $cos x = 3 / 5$ , calculate $cos 2 x$ .
2 marks
If $tan x = 1 / 2$ , calculate $tan 2 x$ .
2 marks
Rewrite ${cos}^{2} (3 x)$ using a multiple-angle cosine.
2 marks
Find $cos (π / 8)$ exactly.
2 marks
Solve $cos 2 x = 1$ on $[0, 2 π)$ .
2 marks
For $x$ in quadrant II with $sin x = 3 / 5$ , find $sin 2 x$ .
2 marks

Read the worked solutions

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

If $cos x = 3 / 5$ , calculate $cos 2 x$ .
2 marks
$2 (9 / 25) - 1 = - 7 / 25$ .
$- 7 / 25$ .
If $tan x = 1 / 2$ , calculate $tan 2 x$ .
2 marks
$2 (1 / 2) / (1 - 1 / 4) = 4 / 3$ .
$4 / 3$ .
Rewrite ${cos}^{2} (3 x)$ using a multiple-angle cosine.
2 marks
Apply the square identity with angle $3 x$ .
$(1 + cos 6 x) / 2$ .
Find $cos (π / 8)$ exactly.
2 marks
Use $(1 + cos (π / 4)) / 2$ for its square, then the positive root.
$\sqrt{2 + \sqrt{2}} / 2$ .
Solve $cos 2 x = 1$ on $[0, 2 π)$ .
2 marks
$2 x = 2 k π$ gives $x = k π$ .
$x = 0, π$ .
For $x$ in quadrant II with $sin x = 3 / 5$ , find $sin 2 x$ .
2 marks
Cosine is $- 4 / 5$ ; multiply $2 (3 / 5) (- 4 / 5)$ .
$- 24 / 25$ .

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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 ↗

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