This note covers the Pythagorean identity, the identity for tangent, and the double angle identities for sine and cosine. You will use them to simplify expressions, prove statements, find exact values and, in a later note, to solve trigonometric equations that mix different angles.
Key ideas
An identity is a statement that is true for every value of the variable where both sides are defined. This is different from an equation, which is true only for particular values. The symbol \(\equiv\) is sometimes used for an identity.
- The Pythagorean identity comes from the unit circle, where a point \((\cos\theta, \sin\theta)\) is at distance 1 from the origin: \(\sin^2\theta + \cos^2\theta = 1\). Here \(\sin^2\theta\) means \((\sin\theta)^2\).
- Rearranging gives two useful forms: \(\sin^2\theta = 1 - \cos^2\theta\) and \(\cos^2\theta = 1 - \sin^2\theta\).
- The tangent identity is \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).
- The double angle identity for sine is \(\sin 2\theta = 2\sin\theta\cos\theta\).
- The double angle identity for cosine has three forms: \(\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\). The second and third forms follow from the first and the Pythagorean identity.
- Choose the form of \(\cos 2\theta\) that fits the problem. If the problem has only \(\sin\theta\) in it, use \(1 - 2\sin^2\theta\). If it has only \(\cos\theta\), use \(2\cos^2\theta - 1\).
How to prove an identity. Start with the more complicated side and change it step by step until it matches the other side. Do not work on both sides at once and do not treat it as an equation to solve. Useful moves are: replace \(\tan\) by \(\frac{\sin}{\cos}\), replace \(1\) by \(\sin^2 + \cos^2\), replace a double angle by its expansion, factorise, or write over a common denominator.
Finding one ratio from another. If you are given one of \(\sin\theta\) or \(\cos\theta\), use the Pythagorean identity to find the other. You get a \(\pm\) result, so use the quadrant given in the question to choose the sign.
Turning one form into another. To get \(2\cos^2\theta - 1\) from \(\cos^2\theta - \sin^2\theta\), replace \(\sin^2\theta\) with \(1 - \cos^2\theta\). To get \(1 - 2\sin^2\theta\), replace \(\cos^2\theta\) with \(1 - \sin^2\theta\) instead. These two rearrangements are also useful in reverse: \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\) and \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\).
Spotting when to use a double angle. If an expression contains both \(\theta\) and \(2\theta\), rewrite the double angle so that only \(\theta\) remains. If the expression is a product \(\sin\theta\cos\theta\), you can reverse the rule to get \(\frac{1}{2}\sin 2\theta\). For instance, \(6\sin x\cos x = 3\sin 2x\).
These identities also let you turn some equations into quadratics in one trigonometric function, which is the method used when solving trigonometric equations.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\sin^2\theta + \cos^2\theta = 1\) | Links sine and cosine. Use it to swap \(\sin^2\) for \(\cos^2\) or find one ratio from the other. |
| \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) | Rewrites tangent so it can be combined with other terms. |
| \(\sin 2\theta = 2\sin\theta\cos\theta\) | Expands a double angle sine into single angle terms. |
| \(\cos 2\theta = \cos^2\theta - \sin^2\theta\) | Basic form of the cosine double angle. |
| \(\cos 2\theta = 2\cos^2\theta - 1\) | Use when the problem is in terms of \(\cos\theta\) only. |
| \(\cos 2\theta = 1 - 2\sin^2\theta\) | Use when the problem is in terms of \(\sin\theta\) only. |
Worked examples
Example 1 The angle \(\theta\) is obtuse and \(\sin\theta = \dfrac{5}{13}\). Find the exact values of \(\sin 2\theta\) and \(\cos 2\theta\).
First find \(\cos\theta\): \(\cos^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}\). An obtuse angle has a negative cosine, so \(\cos\theta = -\dfrac{12}{13}\).
Sine double angle: \(\sin 2\theta = 2\sin\theta\cos\theta = 2 \times \dfrac{5}{13} \times \left(-\dfrac{12}{13}\right) = -\dfrac{120}{169}\).
Cosine double angle, using the sine-only form: \(\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2 \times \dfrac{25}{169} = \dfrac{169 - 50}{169} = \dfrac{119}{169}\).
Check: \(\sin^2 2\theta + \cos^2 2\theta = \dfrac{14400 + 14161}{28561} = \dfrac{28561}{28561} = 1\).
Answer: \(\sin 2\theta = -\frac{120}{169}\), \(\cos 2\theta = \frac{119}{169}\).
Example 2 Prove that \(\dfrac{1 - \cos 2\theta}{\sin 2\theta} \equiv \tan\theta\).
Start with the left side. Use \(1 - 2\sin^2\theta\) for the cosine and \(2\sin\theta\cos\theta\) for the sine.
\(\text{LHS} = \dfrac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} = \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta}\).
Cancel the common factor \(2\sin\theta\): \(\text{LHS} = \dfrac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}\).
Answer: the left side simplifies to \(\tan\theta\), so the identity is proved.
Example 3 Use a double angle identity to show that \(\sin\dfrac{\pi}{8} = \dfrac{\sqrt{2 - \sqrt{2}}}{2}\).
Use \(\cos 2\theta = 1 - 2\sin^2\theta\) with \(\theta = \dfrac{\pi}{8}\), so \(2\theta = \dfrac{\pi}{4}\).
\(\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2} = 1 - 2\sin^2\dfrac{\pi}{8}\).
Rearrange: \(\sin^2\dfrac{\pi}{8} = \dfrac{1 - \frac{\sqrt{2}}{2}}{2} = \dfrac{2 - \sqrt{2}}{4}\).
The angle \(\frac{\pi}{8}\) is acute, so its sine is positive: \(\sin\dfrac{\pi}{8} = \dfrac{\sqrt{2 - \sqrt{2}}}{2} \approx 0.383\).
Answer: \(\sin\frac{\pi}{8} = \frac{\sqrt{2 - \sqrt{2}}}{2} \approx 0.383\).
Common errors
- Writing \(\sin 2\theta = 2\sin\theta\). The correct identity is \(\sin 2\theta = 2\sin\theta\cos\theta\). For \(\theta = 30^\circ\), \(\sin 60^\circ \approx 0.866\) but \(2\sin 30^\circ = 1\).
- Writing \(\cos 2\theta = 2\cos\theta\) or \(\cos 2\theta = \cos^2\theta\). Always use one of the three correct forms.
- Losing the \(\pm\) when taking a square root, or keeping both signs when the quadrant fixes the sign.
- When proving an identity, moving terms across the equals sign or squaring both sides. Work on one side only until it matches the other.
- Cancelling a term instead of a factor, for example cancelling \(\sin\theta\) from \(\frac{1 + \sin\theta}{\sin\theta}\).
Exam tip: In a "show that" or "prove" question, write the identity you use at each step, such as "using \(\cos 2\theta = 1 - 2\sin^2\theta\)", and finish with the exact expression you were asked for.
Check yourself
- Given \(\sin\theta = \dfrac{4}{5}\) with \(\theta\) acute, find \(\sin 2\theta\) and \(\cos 2\theta\).
Answer
\(\cos\theta = \frac{3}{5}\) (positive for an acute angle). \(\sin 2\theta = 2\cdot\frac{4}{5}\cdot\frac{3}{5} = \frac{24}{25}\). \(\cos 2\theta = 1 - 2\cdot\frac{16}{25} = -\frac{7}{25}\). - Simplify \(\dfrac{\sin 2x}{2\sin x}\).
Answer
\(\frac{2\sin x\cos x}{2\sin x} = \cos x\), for \(\sin x \neq 0\). - Given \(\sin x = \dfrac{1}{3}\), find \(\cos 2x\).
Answer
\(\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{1}{9} = \frac{7}{9}\). - Show that \((\cos x + \sin x)^2 \equiv 1 + \sin 2x\).
Answer
Expand: \(\cos^2 x + 2\sin x\cos x + \sin^2 x\). Use \(\sin^2 x + \cos^2 x = 1\) and \(2\sin x\cos x = \sin 2x\) to get \(1 + \sin 2x\).