Reduction Formulae and Co-functions

Reduction formulae let you rewrite the trig ratio of any angle as a ratio of an acute angle. They are the starting point of almost every Paper 2 trigonometry question, especially the "without a calculator" simplifications worth 5 to 7 marks.

Reduction formulae

Use the CAST diagram to find the sign of each ratio in each quadrant: All positive in quadrant 1, Sin in quadrant 2, Tan in quadrant 3, Cos in quadrant 4.

For angles of the form $180° \pm θ$ and $360° \pm θ$ , the ratio stays the same and only the sign changes:

$sin (180° - θ) = sin θ$ , $cos (180° - θ) = - cos θ$ , $tan (180° - θ) = - tan θ$
$sin (180° + θ) = - sin θ$ , $cos (180° + θ) = - cos θ$ , $tan (180° + θ) = tan θ$
$sin (360° - θ) = - sin θ$ , $cos (360° - θ) = cos θ$ , $tan (360° - θ) = - tan θ$

For angles larger than 360°, subtract multiples of 360° first: $sin 480° = sin 120° = sin 60°$ .

Example. $cos 210° = cos (180° + 30°) = - cos 30° = - \frac{3}{2}$

Co-functions

For angles of the form $90° \pm θ$ , sine and cosine swap:

$sin (90° - θ) = cos θ$ and $cos (90° - θ) = sin θ$
$sin (90° + θ) = cos θ$ and $cos (90° + θ) = - sin θ$

The sign comes from the quadrant: $90° + θ$ lies in quadrant 2, where cosine is negative.

Example. $cos 56° = cos (90° - 34°) = sin 34°$

Co-functions often hide in questions: if you see $sin 20°$ and $cos 70°$ together, they are equal.

Negative angles

A negative angle is measured clockwise. Treat $- θ$ as $360° - θ$ :

$sin (- θ) = - sin θ$
$cos (- θ) = cos θ$
$tan (- θ) = - tan θ$

Example. $sin (- 150°) = - sin 150° = - sin 30° = - \frac{1}{2}$

When you have $sin (θ - 180°)$ , write it as $sin (- (180° - θ)) = - sin (180° - θ) = - sin θ$ .

Special angles without a calculator

Learn these values. They come from the 30°-60°-90° triangle with sides $1$ , $3$ , $2$ and the 45° triangle with sides $1$ , $1$ , $2$ :

$sin 30° = \frac{1}{2}$ , $cos 30° = \frac{3}{2}$ , $tan 30° = \frac{1}{3}$
$sin 45° = cos 45° = \frac{1}{2}$ , $tan 45° = 1$
$sin 60° = \frac{3}{2}$ , $cos 60° = \frac{1}{2}$ , $tan 60° = 3$
$sin 0° = 0$ , $cos 0° = 1$ , $sin 90° = 1$ , $cos 90° = 0$

Example. $\frac{cos 300° \cdot tan 225°}{sin 150°} = \frac{\frac{1}{2} \cdot 1}{\frac{1}{2}} = 1$

If a ratio is given as a variable, such as $sin 34° = p$ , draw a right-angled triangle with opposite side $p$ and hypotenuse 1. The adjacent side is $1 - p^{2}$ .

Key terms

CAST diagram: A memory aid showing which ratios are positive in each quadrant.
Reduction formula: A rule that rewrites the ratio of $180° \pm θ$ , $360° \pm θ$ or $- θ$ in terms of $θ$ .
Co-function: Sine and cosine are co-functions: $sin (90° - θ) = cos θ$ .
Special angles: 30°, 45° and 60°, whose ratios have exact values.

Common mistakes

Swapping sine and cosine for $180° \pm θ$ . They only swap for $90° \pm θ$ .
Getting the sign from the ratio of $θ$ instead of from the quadrant of the original angle.
Writing $cos (- θ) = - cos θ$ . Cosine of a negative angle stays positive.
Leaving surds in the denominator when the memo expects a simplified answer.
Using a calculator when the question says not to. Without the exact working you lose most of the marks.

Practice questions

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

Question 1

4 marks total

knowledge

Simplify fully: $\frac{sin ( 180° - x ) \cdot cos ( - x )}{cos ( 90° + x )}$

Worked solution

Hidden - try the question first

Solution

$sin (180° - x) = sin x$
$cos (- x) = cos x$
$cos (90° + x) = - sin x$
$\frac{sin x cos x}{- sin x} = - cos x$

Question 2

5 marks total

routine

Without using a calculator, determine the value of $\frac{cos 300° \cdot tan 225°}{sin 150°}$ .

Worked solution

Hidden - try the question first

Solution

$sin (180° - x) = sin x$
$tan 225° = tan 45° = 1$
$sin 150° = sin 30° = \frac{1}{2}$
$\frac{\frac{1}{2} \times 1}{\frac{1}{2}}$
$= 1$

Question 3

6 marks total

complex

If $sin 34° = p$ , express each of the following in terms of $p$ :

(a) $sin 214°$ (1)

(b) $cos 56°$ (1)

(d) $tan (- 34°)$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $sin 214° = sin (180° + 34°) = - sin 34° = - p$
(b) $cos 56° = cos (90° - 34°) = sin 34° = p$
(c) Triangle with opposite $p$ and hypotenuse 1, so adjacent $= 1 - p^{2}$
$cos 34° = 1 - p^{2}$
(d) $tan (- 34°) = - tan 34°$
$= - \frac{p}{1 - p ^{2}}$

Question 4

8 marks total

problem-solving

Without using a calculator, determine the value of

$sin^{2} 1° + sin^{2} 2° + sin^{2} 3° + \dots + sin^{2} 88° + sin^{2} 89°$

Worked solution

Hidden - try the question first

Solution

Use co-functions: $sin 214° = sin (180° + 34°) = - sin 34° = - p$ , $sin 88° = cos 2°$ , and so on
Pair the terms: $sin^{2} 1° + sin^{2} 89° = sin^{2} 1° + cos^{2} 1° = 1$
There are 44 such pairs, from $(1°, 89°)$ to $(44°, 46°)$ , giving 44
The middle term is $sin^{2} 45° = (\frac{1}{2})^{2} = \frac{1}{2}$
Total $= 44 + \frac{1}{2} = 44, 5$

FAQs

How do I remember the reduction formulae?

Do not memorise a list. Decide which quadrant the angle is in, use CAST for the sign, and only swap sine and cosine for $90° \pm θ$ .

Can I use a calculator for reduction questions?

Only if the question allows it. Most reduction questions say 'without a calculator'. There the marks are for the exact working, so an answer read off a calculator earns little or no credit.

Are reciprocal ratios like cosec and sec examined?

You may use them, but CAPS does not test them explicitly. Most memos use sine, cosine and tangent only.