3.6 Trigonometric identities (IB Maths: Analysis and Approaches SL): Revision notes

Section 1

The Pythagorean identity

For any angle $\theta$ , the point $(\cos\theta, \sin\theta)$ lies on the unit circle $x^2 + y^2 = 1$ , which gives the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1.$ It holds for every angle, in degrees or radians. Rearranged, $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$ .

Notation: $\sin^2\theta$ means $(\sin\theta)^2$ , not $\sin(\theta^2)$ .

Key termsPythagorean identity

Common mistake

$\sin^2\theta = 1 - \cos^2\theta$ does NOT mean $\sin\theta = 1 - \cos\theta$ . Square roots do not split over subtraction.

Section 2

Finding one ratio from another

Given one ratio you can find the others without finding the angle:

Use $\cos^2\theta + \sin^2\theta = 1$ (or a right-angled triangle) to get the size of the other ratio.
Use the quadrant to choose the sign: in the first quadrant all ratios are positive; second quadrant only $\sin$ ; third quadrant only $\tan$ ; fourth quadrant only $\cos$ .
Use $\tan\theta = \frac{\sin\theta}{\cos\theta}$ .

Example: $\theta$ obtuse, $\sin\theta = \frac{5}{13}$ . Then $\cos^2\theta = \frac{144}{169}$ and, as $\theta$ is in the second quadrant, $\cos\theta = -\frac{12}{13}$ and $\tan\theta = -\frac{5}{12}$ .

If no quadrant is given, there are two possible signs, so give both possible values.

Key termsquadrant

Exam tip

When $\tan\theta = \frac{p}{q}$ and $\theta$ is acute, sketch a mental right-angled triangle with sides $p$ and $q$ ; the hypotenuse is $\sqrt{p^2 + q^2}$ .

Common mistake

Taking the positive square root out of habit. Always decide the sign from the quadrant and say why.

Section 3

Double angle identities

The double angle identities (in the formula booklet) are $\sin 2\theta = 2\sin\theta\cos\theta,$ $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta.$ Choose the form of $\cos 2\theta$ that uses the ratio you already know: if you know $\sin\theta$ , use $1 - 2\sin^2\theta$ ; if you know $\cos\theta$ , use $2\cos^2\theta - 1$ .

Example: $\cos x = \frac{3}{4}$ , $x$ acute. Then $\sin x = \frac{\sqrt{7}}{4}$ and $\sin 2x = 2\cdot\frac{\sqrt{7}}{4}\cdot\frac{3}{4} = \frac{3\sqrt{7}}{8}$ .

Key termsdouble angle identities

Common mistake

$\sin 2\theta \ne 2\sin\theta$ and $\cos 2\theta \ne 2\cos\theta$ . Test with $\theta = \frac{\pi}{2}$ : $\sin\pi = 0$ but $2\sin\frac{\pi}{2} = 2$ .

Exam tip

The sign of $\sin 2\theta$ depends on where $2\theta$ lies, not where $\theta$ lies. For obtuse $\theta$ , $2\theta$ is between $\pi$ and $2\pi$ , so $\sin 2\theta < 0$ .

Section 4

Using identities in 'show that' questions

To prove an identity, start from one side (usually the more complicated one) and transform it into the other, one justified step at a time. Useful moves:

Replace $\sin 2x$ by $2\sin x\cos x$ .
Pick the form of $\cos 2x$ that cancels a constant: $1 + \cos 2x = 2\cos^2 x$ and $1 - \cos 2x = 2\sin^2 x$ .
Replace $\frac{\sin x}{\cos x}$ by $\tan x$ .

Example: $\frac{\sin 2x}{1 + \cos 2x} = \frac{2\sin x\cos x}{2\cos^2 x} = \tan x$ .

Key termsidentityshow that

Common mistake

In a 'show that' question, do not start from the answer and do the same thing to both sides as if it were an equation you are solving; that assumes what you are proving.

Must know

$\cos^2\theta + \sin^2\theta = 1$ for all $\theta$ .
$\tan\theta = \frac{\sin\theta}{\cos\theta}$ .
$\sin 2\theta = 2\sin\theta\cos\theta$ ; $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$ .
Signs come from the quadrant: justify every choice of $\pm$ .
You can find $\sin 2x$ , $\cos 2x$ or $\tan x$ from one ratio without ever finding $x$ .

That's the notes covered.

Carry on to the next subtopic.