Trigonometry

Secant, cosecant and cotangent: three reciprocals, not three inverses

Cambridge writes the three reciprocal functions in a fixed way, and you should use these exact definitions: $\sec \theta =\frac{1}{\cos \theta }$, $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$ and $\cot \theta =\frac{1}{\tan \theta }=\frac{\cos \theta }{\sin \theta }$. The second form of $\cot \theta$ is the one that does the work in equations, because it turns a cotangent into a sine and a cosine that you can multiply by.

These are reciprocals, not inverses. $\sec \theta$ is not ${\cos }^{-1}\theta$: the inverse function answers 'which angle has this cosine', while the reciprocal answers 'what is 1 divided by this cosine'. $\sec {60}^{\circ }=2$, but ${\cos }^{-1}2$ has no value at all. Writing ${\sec }^{-1}$ on a script is a common way to lose marks, so avoid the notation entirely.

Each function is undefined wherever its partner is zero. $\sec \theta$ has no value where $\cos \theta =0$, that is at $\theta ={90}^{\circ },{270}^{\circ },\dots$; $\mathrm{cosec}\theta$ has no value where $\sin \theta =0$, that is at $0^{\circ },{180}^{\circ },{360}^{\circ },\dots$; and $\cot \theta$ has no value at those same multiples of ${180}^{\circ }$. A question that asks for the range of a function, or where its graph has an asymptote, is really asking where the denominator vanishes.

The graphs and the properties Cambridge tests

Two functions have period ${180}^{\circ }$ and four have period ${360}^{\circ }$. $\tan \theta$ and $\cot \theta$ repeat every ${180}^{\circ }$; $\sin \theta$, $\cos \theta$, $\sec \theta$ and $\mathrm{cosec}\theta$ repeat every ${360}^{\circ }$. That single fact decides how many solutions an equation has in a given range, so it is worth knowing rather than looking up.

The graphs of $\sec \theta$ and $\mathrm{cosec}\theta$ are not curves you can sketch from memory unless you build them from $\cos \theta$ and $\sin \theta$. Draw the cosine curve first, then put a branch of the secant curve through each arch: where $\cos \theta =1$ the secant graph touches $y=1$, where $\cos \theta$ falls to zero the secant graph shoots off to an asymptote, and in between it dips to a minimum at the point where $\cos \theta$ is largest. The cosecant curve is the same construction on the sine curve, turned sideways by ${90}^{\circ }$.

For angles of any magnitude you work from the related acute angle and then fix the sign. $\sec {210}^{\circ }$ is $\frac{1}{\cos {210}^{\circ }}$, and $\cos {210}^{\circ }=-\cos {30}^{\circ }=-\frac{\sqrt{3}}{2}$, so $\sec {210}^{\circ }=-\frac{2}{\sqrt{3}}$. The sign is the part students drop, so write the quadrant down before you write the number.

• $\sin (-\theta )=-\sin \theta$, $\tan (-\theta )=-\tan \theta$, $\mathrm{cosec}(-\theta )=-\mathrm{cosec}\theta$: these functions are odd
• $\cos (-\theta )=\cos \theta$ and $\sec (-\theta )=\sec \theta$: these two are even
• $\cot \theta$ is decreasing everywhere it is defined, because its gradient is $-{\mathrm{cosec}}^2\theta$, which is always negative
• The asymptotes of $y=\sec \theta$ are the lines $\theta ={90}^{\circ }+{180}^{\circ }n$; those of $y=\mathrm{cosec}\theta$ and $y=\cot \theta$ are $\theta ={180}^{\circ }n$

The two Pythagorean identities

Everything new in this section comes from dividing ${\sin }^2\theta +{\cos }^2\theta =1$ by one of its two squares. Divide every term by ${\cos }^2\theta$ and you get ${\tan }^2\theta +1={\sec }^2\theta$. Divide every term by ${\sin }^2\theta$ and you get $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$. Those are the only two new identities on the syllabus, and Cambridge writes them in the order ${\sec }^2\theta =1+{\tan }^2\theta$ and ${\mathrm{cosec}}^2\theta =1+{\cot }^2\theta$.

They are used in three ways. The first is exact evaluation: if $\tan \theta =\frac{3}{4}$ and $\theta$ is acute, then ${\sec }^2\theta =1+\frac{9}{16}=\frac{25}{16}$, so $\sec \theta =\frac{5}{4}$ and $\cos \theta =\frac{4}{5}$. If $\theta$ were obtuse instead, $\sec \theta$ would be $-\frac{5}{4}$ and $\cos \theta$ would be $-\frac{4}{5}$: the identity gives you the size, the quadrant gives you the sign.

The second use is turning an equation into a quadratic. A question containing ${\sec }^2\theta$ and $\tan \theta$ is a quadratic in $\tan \theta$ in disguise, and a question containing ${\mathrm{cosec}}^2\theta$ and $\cot \theta$ is a quadratic in $\cot \theta$. The third use is proving an identity: replacing ${\sec }^2\theta -1$ by ${\tan }^2\theta$, or ${\mathrm{cosec}}^2\theta -{\cot }^2\theta$ by 1, collapses most of the expressions Cambridge sets.

Exact values of all six functions

The exact values you already know for $\sin$, $\cos$ and $\tan$ at $0^{\circ },{30}^{\circ },{45}^{\circ },{60}^{\circ }$ and ${90}^{\circ }$ give you exact values for the other three by taking reciprocals, and Cambridge expects the surd forms rather than decimals. $\sec {45}^{\circ }=\sqrt{2}$, $\mathrm{cosec}{30}^{\circ }=2$, $\cot {60}^{\circ }=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$. If a value is 'undefined', say so in words: $\cot 0^{\circ }$ is undefined because $\sin 0^{\circ }=0$.

Learn the table below rather than deriving it in the exam, but check any value you are unsure of against the rule that it is the reciprocal of the sine, cosine or tangent in the same column. A quick sanity check is that $\sec \theta$ and $\mathrm{cosec}\theta$ are never between $-1$ and $1$, so $\sec {60}^{\circ }=\frac{1}{2}$ is impossible on sight.

Compound angle formulae

The three expansions are $\sin (A+B)=\sin A\cos B+\cos A\sin B$, $\cos (A+B)=\cos A\cos B-\sin A\sin B$ and $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. For the difference, the sine formula keeps its minus, the cosine formula changes sign, and the tangent formula changes the sign in the numerator: $\sin (A-B)=\sin A\cos B-\cos A\sin B$, $\cos (A-B)=\cos A\cos B+\sin A\sin B$.

Watch the signs in the cosine formula. It is the one that does the opposite of what you expect: a plus inside the bracket gives a minus in the expansion. Every year this is worth a mark, and it is also the reason $\cos 2A={\cos }^{2}A-{\sin }^{2}A$ has a minus in it while $\sin 2A$ does not.

The two jobs these formulae do are exact values and simplification. For an exact value, split the angle into two you know: ${75}^{\circ }={45}^{\circ }+{30}^{\circ }$, so $\sin {75}^{\circ }=\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times \frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$. For a simplification, expand both terms and collect like terms: $\cos (x+{45}^{\circ })-\cos (x-{45}^{\circ })=\frac{\sqrt{2}}{2}(\cos x-\sin x)-\frac{\sqrt{2}}{2}(\cos x+\sin x)=-\sqrt{2}\sin x$, which is the shape of answer Cambridge wants.

• $\sin (A+B)=\sin A\cos B+\cos A\sin B$; $\sin (A-B)=\sin A\cos B-\cos A\sin B$
• $\cos (A+B)=\cos A\cos B-\sin A\sin B$; $\cos (A-B)=\cos A\cos B+\sin A\sin B$
• $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$; replace every $+$ by $-$ for $\tan (A-B)$
• Useful splits: ${75}^{\circ }={45}^{\circ }+{30}^{\circ }$, ${15}^{\circ }={45}^{\circ }-{30}^{\circ }$, ${105}^{\circ }={60}^{\circ }+{45}^{\circ }$

Double angle formulae

Putting $B=A$ into the compound angle formulae gives $\sin 2A=2\sin A\cos A$, $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$ and $\cos 2A={\cos }^{2}A-{\sin }^{2}A$. The cosine version is the important one, because it has two more forms obtained from ${\sin }^{2}A+{\cos }^{2}A=1$: $\cos 2A=2{\cos }^{2}A-1$ and $\cos 2A=1-2{\sin }^{2}A$. Choose whichever of the three leaves you with a single trigonometric function.

That choice is the whole skill. An equation containing $\cos 2\theta$ and $\cos \theta$ wants the form $2{\cos }^2\theta -1$, because it turns into a quadratic in $\cos \theta$. An equation containing $\cos 2\theta$ and $\sin \theta$ wants $1-2{\sin }^2\theta$ for the same reason. An equation containing $\cos 2\theta$ and $\sin \theta \cos \theta$ wants ${\cos }^2\theta -{\sin }^2\theta$. Using the wrong form leaves you with two different functions and nowhere to go.

The formulae also run backwards, which is how they are used to evaluate something exactly. Given $\tan A=\frac{1}{3}$ with $A$ acute, $\tan 2A=\frac{2/3}{1-1/9}=\frac{2/3}{8/9}=\frac{3}{4}$. Given $\cos \theta =\frac{4}{5}$, $\sin 2\theta =2\times \frac{3}{5}\times \frac{4}{5}=\frac{24}{25}$, using $\sin \theta =\frac{3}{5}$ from the $3,4,5$ triangle.

Solving equations: the three shapes Cambridge sets

Shape 1: a quadratic in one function. $\tan \theta +\cot \theta =4$ looks like it needs two functions, but $\cot \theta =\frac{1}{\tan \theta }$, so multiplying by $\tan \theta$ gives ${\tan }^2\theta -4\tan \theta +1=0$, which the quadratic formula solves as $\tan \theta =2\pm \sqrt{3}$. ${\sec }^2\theta -\tan \theta =5$ is the same idea one step earlier: replace ${\sec }^2\theta$ by $1+{\tan }^2\theta$ and you have ${\tan }^2\theta -\tan \theta -4=0$. Both then need every angle in the range, so find one angle with your calculator and use the period of $\tan$, which is ${180}^{\circ }$.

Shape 2: one factor of the whole equation. $2\sin 2\theta =\cos \theta$ becomes $4\sin \theta \cos \theta =\cos \theta$, and the temptation is to divide by $\cos \theta$ and lose solutions. Move everything to one side instead: $\cos \theta (4\sin \theta -1)=0$. You then have two separate equations, $\cos \theta =0$ and $\sin \theta =\frac{1}{4}$, and between them they give every solution. Dividing by $\cos \theta$ throws away the ${90}^{\circ }$ and ${270}^{\circ }$ answers, which are usually worth two of the marks.

Shape 3: a sine and a cosine added together. $a\sin \theta +b\cos \theta =c$ will not factorise, so it has to be written as a single $R\sin (\theta +\alpha )$ first. If the question is in $2\theta$, double the range before you start: for $0^{\circ }\le \theta \le {360}^{\circ }$ you need all values of $2\theta$ from $0^{\circ }$ to ${720}^{\circ }$, and each one you find has to be halved at the end.

• One equation, one function: use an identity until only $\sin \theta$, $\cos \theta$ or $\tan \theta$ is left
• Never divide by $\cos \theta$ or $\sin \theta$: factorise instead, or you lose the solutions where that factor is zero
• After solving, list the solutions in order and check each one against the range the question gave you
• If the equation is in $2\theta$, $3\theta$ or $\theta +{30}^{\circ }$, work in the new variable, then convert back at the end

Writing a sin theta + b cos theta in the form R sin(theta + alpha)

Expand $R\sin (\theta +\alpha )=R\sin \theta \cos \alpha +R\cos \theta \sin \alpha$ and match it against $a\sin \theta +b\cos \theta$. Matching coefficients gives $R\cos \alpha =a$ and $R\sin \alpha =b$, so $R=\sqrt{a^2+b^2}$ and $\tan \alpha =\frac{b}{a}$. Use $R\sin (\theta -\alpha )$ when the cosine term is negative, and the cosine form when the question asks for it: $R\cos (\theta -\alpha )=R\cos \theta \cos \alpha +R\sin \theta \sin \alpha$, which matches $a\cos \theta +b\sin \theta$ with $R\cos \alpha =a$ and $R\sin \alpha =b$.

For $3\sin \theta -4\cos \theta$: $R=\sqrt{9+16}=5$ and $\tan \alpha =\frac{4}{3}$ with $\alpha ={53.13}^{\circ }$, so the expression is $5\sin (\theta -{53.13}^{\circ })$. The point of the form is that the maximum and minimum are now visible: the largest $\sin$ can be is 1, so the expression lies between $-5$ and $5$. The maximum is 5, and it happens when $\theta -{53.13}^{\circ }={90}^{\circ }$, that is $\theta ={143.13}^{\circ }$.

Two checks keep the sign of $\alpha$ honest. First, $\alpha$ is always acute, between $0^{\circ }$ and ${90}^{\circ }$, because it comes from $\tan \alpha =\frac{|b|}{|a|}$; the signs of $a$ and $b$ decide whether the bracket contains $+\alpha$ or $-\alpha$. Second, test your answer at $\theta =0$. For $3\sin \theta -4\cos \theta$ the original gives $-4$, and the form gives $5\sin (-{53.13}^{\circ })=-4$. If those disagree, you have chosen the wrong sign inside the bracket.

Angles of any magnitude: getting all the solutions

Your calculator gives one angle for an inverse sine, cosine or tangent. That angle is never the whole answer: a trigonometric equation has a second solution in each full turn of the circle, and possibly more. For $\sin \theta =k$, the second solution in the range is ${180}^{\circ }-{\theta }_1$; for $\cos \theta =k$, it is ${360}^{\circ }-{\theta }_1$; for $\tan \theta =k$, it is ${\theta }_1+{180}^{\circ }$. Then add or subtract ${360}^{\circ }$ (or ${180}^{\circ }$ for tangent) until you leave the range.

The cleanest way to write this up is to find one value, add the period repeatedly, and check each candidate against the range. For $\tan \theta =2.5616$, the calculator gives ${68.66}^{\circ }$, and adding ${180}^{\circ }$ gives ${248.66}^{\circ }$; for $\tan \theta =-1.5616$, the calculator gives $-{57.34}^{\circ }$, so add ${180}^{\circ }$ to get ${122.66}^{\circ }$ and again to get ${302.66}^{\circ }$. That is the full set of four, and it is the adding-the-period step, not the calculator step, that carries the marks.

Round at the end, not in the middle. If a question asks for answers correct to one decimal place, keep full accuracy through the working and round only the final angles; rounding ${53.13}^{\circ }$ to ${53.1}^{\circ }$ before you subtract it from ${90}^{\circ }$ is the sort of slip that turns a correct method into a wrong answer.