Trigonometric identities and reduction to an acute angle – Fundamentals of Trigonometry (Senior 4 Mathematics) | Ganzaa

Fundamentals of Trigonometry

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Senior 4 · Mathematics

Fundamentals of Trigonometry

5 lessons

Lesson 3 of 5

Trigonometric identities and reduction to an acute angle

25 min

Two shortcuts that save every calculation

Before calculators, trigonometric tables printed values only for angles from $0^\circ$ to $90^\circ$ . That was enough, because the symmetry of the unit circle turns the sine of any angle into plus or minus the sine (or cosine) of an acute angle. And because $M(\cos\theta, \sin\theta)$ lies on a circle of radius 1, the six ratios are tied together by equations that hold for every angle, the identities. Together they let you find any ratio from one known ratio and simplify long expressions.

Definition

Trigonometric identity: an equation involving trigonometric ratios that is true for every value of the variable for which both sides are defined. Example: $\tan x = \frac{\sin x}{\cos x}$ for all $x$ with $\cos x \ne 0$ . An equation such as $\sin x = \frac{1}{2}$ , by contrast, is true only for particular values.

Important

The fundamental identities

\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta .

Proof. $M(\cos\theta, \sin\theta)$ is at distance 1 from $O$ , so by Pythagoras $\cos^2\theta + \sin^2\theta = 1$ . Dividing by $\cos^2\theta$ (where it is not 0) gives $1 + \tan^2\theta = \sec^2\theta$ ; dividing by $\sin^2\theta$ gives $\cot^2\theta + 1 = \csc^2\theta$ . $\blacksquare$

(Here $\sin^2\theta$ means $(\sin\theta)^2$ .) Together with $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and the three reciprocal identities, these are the tools for everything below.

Worked example 1: from one ratio to another

Given $\tan\theta = 2$ and $\pi \lt \theta \lt \frac{3\pi}{2}$ , find $\cos\theta$ and $\sin\theta$ .

Solution $\sec^2\theta = 1 + \tan^2\theta = 5$ , so $\cos^2\theta = \frac{1}{5}$ and $\cos\theta = \pm\frac{1}{\sqrt{5}}$ . In quadrant III cosine is negative: $\cos\theta = -\frac{\sqrt{5}}{5}$ . Then $\sin\theta = \tan\theta\cos\theta = -\frac{2\sqrt{5}}{5}$ . Check: $\frac{4}{5} + \frac{1}{5} = 1$ . ✓

Worked example 2: proving identities

Prove (a) $(1 - \cos^2 x)(1 + \cot^2 x) = 1$ (b) $\sec x - \cos x = \sin x \tan x$ .

Solution Start from the more complicated side and transform it into the other side.

$(1 - \cos^2 x)(1 + \cot^2 x) = \sin^2 x \cdot \csc^2 x = \sin^2 x \cdot \frac{1}{\sin^2 x} = 1$ .
$\sec x - \cos x = \frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x} = \sin x \cdot \frac{\sin x}{\cos x} = \sin x \tan x$ .

Both hold wherever the expressions are defined (in (a), $\sin x \ne 0$ ; in (b), $\cos x \ne 0$ ).

Warning

Do not treat an identity as an equation to solve. Writing both sides and "doing the same thing to both" can hide an error, because you assume what you want to prove. Work on one side only. And remember that $\sin^2 x$ is $(\sin x)^2$ , while $\sin x^2$ means the sine of $x^{2}$ .

Associated angles: reduction to an acute angle

Reflect the point for $\theta$ in the axes and in the origin. The four points have coordinates that differ only in sign:

Angle	Name	$\sin$	$\cos$	$\tan$
$\theta + 2k\pi$	equivalent (coterminal)	$\sin\theta$	$\cos\theta$	$\tan\theta$
$-\theta$	opposite	$-\sin\theta$	$\cos\theta$	$-\tan\theta$
$\pi - \theta$	supplementary	$\sin\theta$	$-\cos\theta$	$-\tan\theta$
$\pi + \theta$	differ by $π \pi$	$-\sin\theta$	$-\cos\theta$	$\tan\theta$
$\frac{\pi}{2} - \theta$	complementary	$\cos\theta$	$\sin\theta$	$\cot\theta$
$\frac{\pi}{2} + \theta$		$\cos\theta$	$-\sin\theta$	$-\cot\theta$

The complementary row comes from reflecting in the line $y = x$ , which swaps the coordinates: in a right triangle, the sine of one acute angle is the cosine of the other ("co-sine" means the sine of the complement).

Method for any angle: (1) remove whole turns; (2) find the quadrant; (3) find the reference angle $\alpha$ , the acute angle between the terminal side and the $x$ -axis; (4) take the ratio of $\alpha$ with the sign given by A, S, T, C.

Worked example 3: exact values of large and negative angles

Find (a) $\sin 150^\circ$ (b) $\cos 225^\circ$ (c) $\tan(-300^\circ)$ (d) $\sin\frac{23\pi}{6}$ .

Solution

Quadrant II, $\alpha = 30^\circ$ , sine positive: $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$ .
Quadrant III, $\alpha = 45^\circ$ , cosine negative: $\cos 225^\circ = -\frac{\sqrt{2}}{2}$ .
$-300^\circ + 360^\circ = 60^\circ$ : $\tan(-300^\circ) = \tan 60^\circ = \sqrt{3}$ .
$\frac{23\pi}{6} - 4\pi = -\frac{\pi}{6}$ , so $\sin\frac{23\pi}{6} = \sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac{1}{2}$ .

Worked example 4: simplifying with associated angles

Simplify $E = \sin(\pi - x) + \cos\left(\frac{\pi}{2} - x\right) + \sin(\pi + x) - \cos(-x)\tan(-x)$ .

Solution $\sin(\pi - x) = \sin x$ , $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ , $\sin(\pi + x) = -\sin x$ , and $\cos(-x)\tan(-x) = \cos x \cdot (-\tan x) = -\sin x$ . So $E = \sin x + \sin x - \sin x + \sin x = 2\sin x$ .

Tip

For angles written as $\pi \pm \theta$ or $2\pi - \theta$ , the ratio keeps its name; for $\frac{\pi}{2} \pm \theta$ or $\frac{3\pi}{2} \pm \theta$ , $\sin$ and $\cos$ swap names. In both cases, fix the sign by imagining $\theta$ small and asking which quadrant the angle is in.

blackpenredpen: Prove the Pythagorean identities. Watch for how 1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ both follow from sin²θ + cos²θ = 1.

✦

Practice

Exact values unless told otherwise.

Given $\cos\theta = 0.6$ and $\frac{3\pi}{2} \lt \theta \lt 2\pi$ , find $\sin\theta$ and $\tan\theta$ .
Prove that $\tan x + \cot x = \sec x \csc x$ .
Prove that $(1 + \tan^2 x)\cos^2 x = 1$ .
Prove that $(\sin x + \cos x)^2 = 1 + 2\sin x\cos x$ .
Prove that $\frac{1}{1 - \sin x} + \frac{1}{1 + \sin x} = 2\sec^2 x$ .
Find $\sin 300^\circ$ , $\cos 135^\circ$ , $\tan 210^\circ$ , $\sin\left(-\frac{7\pi}{6}\right)$ and $\cos\frac{11\pi}{4}$ .
Simplify $\cos(\pi - x) + \cos(\pi + x) + 2\sin\left(\frac{\pi}{2} - x\right)$ .
Simplify $\tan(\pi - x)\tan\left(\frac{\pi}{2} - x\right)$ .

Show answersHide answers

$\sin\theta = -0.8$ (quadrant IV), $\tan\theta = -\frac{4}{3}$ .
$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x} = \sec x\csc x$ .
$\sec^2 x \cdot \cos^2 x = 1$ .
$\sin^2 x + 2\sin x\cos x + \cos^2 x = 1 + 2\sin x\cos x$ .
$\frac{(1 + \sin x) + (1 - \sin x)}{1 - \sin^2 x} = \frac{2}{\cos^2 x} = 2\sec^2 x$ .
$-\frac{\sqrt{3}}{2}$ , $-\frac{\sqrt{2}}{2}$ , $\frac{\sqrt{3}}{3}$ , $\frac{1}{2}$ ( $-\frac{7\pi}{6} + 2\pi = \frac{5\pi}{6}$ ), $-\frac{\sqrt{2}}{2}$ ( $\frac{11\pi}{4} - 2\pi = \frac{3\pi}{4}$ ).
$-\cos x - \cos x + 2\cos x = 0$ .
$(-\tan x)(\cot x) = -1$ .

Quick check

What is the exact value of sin 150°?

Key points

$\sin^2\theta + \cos^2\theta = 1$ , $1 + \tan^2\theta = \sec^2\theta$ , $1 + \cot^2\theta = \csc^2\theta$ .
Prove an identity by transforming one side into the other.
Any angle reduces to its acute reference angle; the sign comes from the quadrant (A, S, T, C).
$\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$ and $\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the tables, the worked examples, the practice items and the questions.
Video: Prove the Pythagorean identities, blackpenredpen, embedded from YouTube.

1Two shortcuts that save every calculation

2Associated angles: reduction to an acute angle

Two shortcuts that save every calculation

Associated angles: reduction to an acute angle