13.11 Trigonometric Identities
SL 3.6
- the identity \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
- the Pythagorean identity \(\cos^2\theta + \sin^2\theta = 1\), and why it is true
- how to find one trigonometric ratio from another without finding the angle
- how to simplify trigonometric expressions and prove identities
Simple trigonometric identities
An identity is an equation that is true for every value of the variable (where both sides are defined). We use the symbol \(\equiv\) when we want to show that an equation is an identity.
Take a point \(P(x, y)\) on a circle with centre \(O\) and radius \(r\), as in Figure 1. Let \(\theta\) be the angle from the positive \(x\)-axis to \(OP\). As in Section 13.6,
\[ \sin\theta = \frac{y}{r}, \qquad \cos\theta = \frac{x}{r}, \qquad \tan\theta = \frac{y}{x} . \]
Now divide \(\sin\theta\) by \(\cos\theta\):
\[ \frac{\sin\theta}{\cos\theta} = \frac{y}{r} \div \frac{x}{r} = \frac{y}{r} \times \frac{r}{x} = \frac{y}{x} = \tan\theta . \]
Identity for \(\tan\theta\) In the formula booklet
\[ \tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cos\theta \neq 0 \]
- \(\theta\): any angle (in degrees or radians) with \(\cos\theta \neq 0\)
Example 1 Given that \(\sin\theta = \dfrac{5}{13}\) and \(\cos\theta = -\dfrac{12}{13}\), find the value of \(\tan\theta\).
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{5}{13} \div \left(-\dfrac{12}{13}\right) = \dfrac{5}{13} \times \left(-\dfrac{13}{12}\right) = -\dfrac{5}{12}\).
(Here \(\sin\theta > 0\) and \(\cos\theta < 0\), so \(\theta\) is in the second quadrant, where \(\tan\) is negative. ✓)
The Pythagorean identity
For the point \(P(x, y)\) in Figure 1, Pythagoras’ theorem gives
\[ x^2 + y^2 = r^2 . \qquad (1) \]
Divide both sides of (1) by \(r^2\):
\[ \left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = 1 \;\Longrightarrow\; \cos^2\theta + \sin^2\theta = 1 . \qquad (2) \]
We write \(\sin^2\theta\) for \((\sin\theta)^2\). It does not mean \(\sin(\theta^2)\).
Pythagorean identity In the formula booklet
\[ \cos^2\theta + \sin^2\theta = 1 \]
- \(\theta\): any angle
- \(\cos^2\theta\), \(\sin^2\theta\): \((\cos\theta)^2\) and \((\sin\theta)^2\)
It is often used in one of the rearranged forms
\[ \sin^2\theta = 1 - \cos^2\theta, \qquad \cos^2\theta = 1 - \sin^2\theta . \]
Finding one ratio from another
The Pythagorean identity lets us find \(\cos\theta\) from \(\sin\theta\) (or the other way) without finding \(\theta\). A square root gives two signs. Use the quadrant of \(\theta\) to choose the sign (see Section 13.6).
Example 2 Given that \(\sin\theta = \dfrac{2}{5}\),
(a) find the possible values of \(\tan\theta\), without finding \(\theta\)
(b) find \(\tan\theta\) when it is also given that \(\dfrac{\pi}{2} < \theta < \pi\).
(a) \(\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{4}{25} = \dfrac{21}{25}\), so \(\cos\theta = \pm\dfrac{\sqrt{21}}{5}\). Then
\[ \tan\theta = \frac{2/5}{\pm\sqrt{21}/5} = \pm\frac{2}{\sqrt{21}} = \pm\frac{2\sqrt{21}}{21} . \]
(b) In the second quadrant \(\cos\theta < 0\), so \(\tan\theta < 0\): \(\;\tan\theta = -\dfrac{2\sqrt{21}}{21}\).
Simplifying expressions
To simplify, it often helps to write everything in terms of \(\sin\) and \(\cos\), and to look for \(1 - \sin^2\theta\) or \(1 - \cos^2\theta\).
Example 3 Simplify \((1 - \sin^2\theta)\tan\theta\).
\((1 - \sin^2\theta)\tan\theta = \cos^2\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta\cos\theta\).
Proving identities
You already know some identities, for example \(\sin^2\theta \equiv 1 - \cos^2\theta\). The left-hand side (LHS) and the right-hand side (RHS) are equal for every value of \(\theta\).
To prove an identity, it is not enough to put a few numbers into both sides and get the same answer. A few numbers can never check every value. We need a general argument.
How to prove an identity
- Start with one side, usually the more complicated one.
- Change it, step by step, using known identities and algebra, until you reach the other side.
- Do not treat it like an equation: do not move terms from one side to the other, and do not do the same thing to both sides.
- End with “\(=\) RHS” (or “\(=\) LHS”).
Useful moves: write in terms of \(\sin\) and \(\cos\); use a common denominator; expand or factorise; replace \(1 - \sin^2\theta\) by \(\cos^2\theta\) (and so on).
Example 4 Prove each identity.
(a) \((\sin x + \cos x)^2 + (\sin x - \cos x)^2 \equiv 2\)
(b) \(\cos^4 x - \sin^4 x \equiv 1 - 2\sin^2 x\)
(a) Expand both brackets:
\[ \begin{aligned} \text{LHS} &= (\sin^2 x + 2\sin x\cos x + \cos^2 x) \\ &\quad + (\sin^2 x - 2\sin x\cos x + \cos^2 x) \\ &= 2(\sin^2 x + \cos^2 x) = 2 = \text{RHS} . \end{aligned} \]
(b) Factorise the difference of two squares:
\[ \begin{aligned} \text{LHS} &= (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) \\ &= \cos^2 x - \sin^2 x \\ &= (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x = \text{RHS} . \end{aligned} \]
HL students: continue with 13.12 Reciprocal Trigonometric Identities.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\sin^2\theta = \sin(\theta^2)\) | \(\sin^2\theta = (\sin\theta)^2\) | The power belongs to the ratio, not to the angle. |
| \(\cos\theta = \sqrt{1 - \sin^2\theta}\) always | \(\cos\theta = \pm\sqrt{1 - \sin^2\theta}\) | Choose the sign from the quadrant of \(\theta\). |
| \(\sin\theta + \cos\theta = 1\) | \(\sin^2\theta + \cos^2\theta = 1\) | The identity is about the squares. Try \(\theta = \frac{\pi}{4}\). |
| “True for \(\theta = 0\) and \(\frac{\pi}{3}\), so it is an identity” | Give a general proof | A few values never prove a statement for all \(\theta\). |
| Doing the same thing to both sides of an identity | Work on one side only | You must not assume what you are trying to prove. |
Exercises
1 Given that \(\cos x = \dfrac{2}{3}\) and \(\dfrac{3\pi}{2} < x < 2\pi\), find the exact values of \(\sin x\) and \(\tan x\).
\(\sin^2 x = 1 - \dfrac{4}{9} = \dfrac{5}{9}\). In the fourth quadrant \(\sin x < 0\), so \(\sin x = -\dfrac{\sqrt{5}}{3}\) and
\[ \tan x = \frac{-\sqrt{5}/3}{2/3} = -\frac{\sqrt{5}}{2} . \]
2 Given that \(\sin\theta = \dfrac{1}{3}\), find the possible values of \(\tan\theta\) without finding \(\theta\).
\(\cos^2\theta = 1 - \dfrac{1}{9} = \dfrac{8}{9}\), so \(\cos\theta = \pm\dfrac{2\sqrt{2}}{3}\) and
\[ \tan\theta = \frac{1/3}{\pm 2\sqrt{2}/3} = \pm\frac{1}{2\sqrt{2}} = \pm\frac{\sqrt{2}}{4} . \]
3 Simplify \(\dfrac{\sin^2 x}{1 - \cos x}\).
\[ \frac{1 - \cos^2 x}{1 - \cos x} = \frac{(1 - \cos x)(1 + \cos x)}{1 - \cos x} = 1 + \cos x . \]
4 Prove that \(\dfrac{1}{1 - \sin x} + \dfrac{1}{1 + \sin x} \equiv \dfrac{2}{\cos^2 x}\).
\[ \begin{aligned} \text{LHS} &= \frac{(1 + \sin x) + (1 - \sin x)}{(1 - \sin x)(1 + \sin x)} \\ &= \frac{2}{1 - \sin^2 x} = \frac{2}{\cos^2 x} = \text{RHS} . \end{aligned} \]