Trigonometric identitiesAQA A-Level Maths: Revision notes

For any angle $\theta$, $\tan\theta\equiv\frac{\sin\theta}{\cos\theta}.$ This is an identity: it is true for every value of $\theta$ for which it is defined. It is undefined when $\cos\theta=0$ (that is, $\theta=90^\circ+180^\circ n$). It also tells you where tangent is zero ($\sin\theta=0$) and why its sign depends on the quadrant. In a right-angled triangle it reproduces $\tan\theta=\frac{\text{opp}}{\text{adj}}$.

For any angle $\theta$, $\sin^2\theta+\cos^2\theta\equiv1.$ It follows from Pythagoras: a point on a circle of radius $1$ has coordinates $(\cos\theta,\sin\theta)$, so $\cos^2\theta+\sin^2\theta=1$. Useful rearrangements:

• $\sin^2\theta=1-\cos^2\theta$
• $\cos^2\theta=1-\sin^2\theta$ Here $\sin^2\theta$ means $(\sin\theta)^2$, not $\sin(\theta^2)$.

If you know one of $\sin\theta$ or $\cos\theta$, the identity gives the other, and then $\tan\theta=\frac{\sin\theta}{\cos\theta}$ gives the third. Example: $\sin\theta=\frac{5}{13}$ with $\theta$ obtuse. Then $\cos^2\theta=1-\frac{25}{169}=\frac{144}{169}$. An obtuse angle has negative cosine, so $\cos\theta=-\frac{12}{13}$ and $\tan\theta=\frac{5/13}{-12/13}=-\frac{5}{12}$. If the problem gives a relationship such as $\sin\theta=2\cos\theta$, substitute it into $\sin^2\theta+\cos^2\theta=1$ to get $5\cos^2\theta=1$, or divide by $\cos\theta$ to get $\tan\theta=2$.

Replace $\tan\theta$ by $\frac{\sin\theta}{\cos\theta}$, then cancel. Examples:

• $\cos x\tan x=\cos x\times\frac{\sin x}{\cos x}=\sin x$
• $\frac{\sin x}{\tan x}=\sin x\times\frac{\cos x}{\sin x}=\cos x$
• $1-\cos^2x=\sin^2x$, and $\frac{\sin^2x}{1-\sin^2x}=\frac{\sin^2x}{\cos^2x}=\tan^2x$ Cancel only factors, never terms: you can cancel $\cos x$ in $\frac{\cos x\sin x}{\cos x}$, but not in $\frac{1+\cos x}{\cos x}$.

To prove an identity, start from one side (usually the more complicated one) and transform it into the other, using valid steps only. Do not work on both sides at once, and do not treat the identity as an equation to rearrange. Worked example: prove $\tan\theta+\frac{1}{\tan\theta}\equiv\frac{1}{\sin\theta\cos\theta}$. $\tan\theta+\frac{1}{\tan\theta}=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}$. Typical strategy: write $\tan$ in terms of $\sin$ and $\cos$, combine fractions over a common denominator, then use $\sin^2\theta+\cos^2\theta=1$ to simplify the numerator.