Double Angle Identities

Q: How do I find sin 2θ if I only know tan θ?

Draw $\theta$ in the correct quadrant, use Pythagoras to find the hypotenuse, read off $\sin\theta$ and $\cos\theta$, then use $2\sin\theta\cos\theta$.

Double angle identities come straight from the compound angle formulae with $α = β$ . Choosing the right form of $cos 2 x$ is the main skill, and it decides whether an identity proof or an equation works out in two lines or ten.

sin 2x and the three forms of cos 2x

$sin 2 x = 2 sin x cos x$

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

The second and third forms come from replacing $cos^{2} x$ or $sin^{2} x$ using $sin^{2} x + cos^{2} x = 1$ .

Choosing a form:

If the rest of the expression has only $sin x$ , use $1 - 2 sin^{2} x$ .
If it has only $cos x$ , use $2 cos^{2} x - 1$ .
If you see $1 + cos 2 x$ , use $2 cos^{2} x - 1$ so the 1s cancel: $1 + cos 2 x = 2 cos^{2} x$ .
If you see $1 - cos 2 x$ , use $1 - 2 sin^{2} x$ : $1 - cos 2 x = 2 sin^{2} x$ .

Example. If $cos x = \frac{1}{3}$ , then $cos 2 x = 2 (\frac{1}{9}) - 1 = - \frac{7}{9}$ .

Simplifying with double angles

Look for double angle patterns hidden in an expression:

$2 sin θ cos θ = sin 2 θ$ , so $sin 15° cos 15° = \frac{1}{2} sin 30° = \frac{1}{4}$
$cos^{2} 22, 5° - sin^{2} 22, 5° = cos 45° = \frac{2}{2}$
$sin 4 x = 2 sin 2 x cos 2 x$ , since $4 x$ is double $2 x$

Example. Simplify $\frac{sin 2 x}{1 + cos 2 x}$ .

$\frac{2 s i n x c o s x}{2 c o s ^{2} x} = \frac{s i n x}{c o s x} = tan x$

Factorise wherever you can. Most simplification questions end with a common factor cancelling.

Key terms

Double angle: An angle of the form $2 x$ .
Square identity: $sin^{2} x + cos^{2} x = 1$ , used to move between the forms of $cos 2 x$ .
Simplify: Rewrite an expression in its shortest equivalent form, often a single ratio.

Common mistakes

Writing $sin 2 x = 2 sin x$ .
Choosing a form of $cos 2 x$ that leaves both $sin x$ and $cos x$ in an equation that needs one ratio.
Cancelling terms instead of factors.
Forgetting that $sin^{2} x$ means $(sin x)^{2}$ , not $sin (x^{2})$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

If $cos x = \frac{1}{3}$ , determine the value of $cos 2 x$ without using a calculator.

Worked solution

Hidden - try the question first

Solution

$cos 2 x = 2 cos^{2} x - 1$
$= 2 (\frac{1}{3})^{2} - 1 = \frac{2}{9} - 1$
$= - \frac{7}{9}$

Question 2

5 marks total

routine

Simplify $\frac{sin 2 x}{1 + cos 2 x}$ to a single trigonometric ratio.

Worked solution

Hidden - try the question first

Solution

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

Question 3

6 marks total

complex

Without using a calculator, determine the value of:

(a) $sin 15° cos 15°$ (3)

(b) $cos^{2} 22, 5° - sin^{2} 22, 5°$ (3)

Worked solution

Hidden - try the question first

Solution

(a) $sin 15° cos 15° = \frac{1}{2} (2 sin 15° cos 15°)$
$= \frac{1}{2} sin 30°$
$= \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$
(b) $= cos (2 \times 22, 5°)$
$= cos 45°$
$= \frac{2}{2}$

Question 4

9 marks total

problem-solving

$tan θ = - \frac{3}{4}$ and $sin θ > 0$ .

(a) Without using a calculator, determine $sin 2 θ$ and $cos 2 θ$ . (4)

(b) Prove that $sin 3 θ = 3 sin θ - 4 sin^{3} θ$ for all $θ$ . (4)

Worked solution

Hidden - try the question first

Solution

(a) $sin 15° cos 15° = \frac{1}{2} (2 sin 15° cos 15°)$ is in quadrant 2, so $sin θ = \frac{3}{5}$ and $cos θ = - \frac{4}{5}$
$sin 2 θ = 2 \cdot \frac{3}{5} \cdot (- \frac{4}{5}) = - \frac{24}{25}$
$cos 2 θ = cos^{2} θ - sin^{2} θ$
$= \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$
(b) $sin 3 θ = sin (2 θ + θ) = sin 2 θ cos θ + cos 2 θ sin θ$
$= 2 sin θ cos^{2} θ + (1 - 2 sin^{2} θ) sin θ$
$= 2 sin θ (1 - sin^{2} θ) + sin θ - 2 sin^{3} θ$
$= 3 sin θ - 4 sin^{3} θ$
(c) $3 \cdot \frac{3}{5} - 4 \cdot \frac{27}{125} = \frac{225 - 108}{125} = \frac{117}{125}$

FAQs

Which form of cos 2x should I use?

Pick the form that matches the rest of the expression. If you need everything in $sin x$ , use $1 - 2 sin^{2} x$ ; for $cos x$ , use $2 cos^{2} x - 1$ .

Are double angle formulae given in the IEB exam?

Yes, they are on the formula sheet, but you need to recognise when and how to use them quickly.

How do I find sin 2θ if I only know tan θ?

Draw $θ$ in the correct quadrant, use Pythagoras to find the hypotenuse, read off $sin θ$ and $cos θ$ , then use $2 sin θ cos θ$ .

sin 2x and the three forms of cos 2x

Simplifying with double angles

Key terms

Common mistakes

Practice questions

FAQs

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