Trigonometry

Everyone uses ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$. Put $x=y=2$.

The left side is $2{\tan }^{-1}2\approx 126.87^{\circ }$. The right side is ${\tan }^{-1}\frac{4}{1-4}={\tan }^{-1}\left(-\frac{4}{3}\right)\approx -53.13^{\circ }$.

The two differ by exactly $180^{\circ }$, and the reason is that ${\tan }^{-1}$ only ever returns a value in $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. The genuine sum here is larger than $\frac{\pi }{2}$, so no arctangent can equal it. The formula is not an identity but a statement that holds when $xy<1$, and otherwise needs a correction:

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}+\{\begin{array}{ll}\pi & x>0,y>0,xy>1\\ -\pi & x<0,y<0,xy>1.\end{array}$

Every inverse trigonometric function carries a range restriction of this kind, and almost every Advanced question on the topic is built on one. The direct functions are periodic and therefore many-to-one; the inverses undo them only on one chosen stretch, and any question whose answer strays outside that stretch behaves differently from what the algebra suggests.

1. The identities everything else is built from

Three Pythagorean relations and two addition formulas generate the whole subject:

${\sin }^2\theta +{\cos }^2\theta =1,\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B,\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B.$

Setting $B=A$ gives the double-angle formulas, and the three forms of $\cos 2A$ — namely $1-2{\sin }^{2}A$, $2{\cos }^{2}A-1$ and $\frac{1-t^2}{1+t^2}$ with $t=\tan A$ — are worth carrying separately, because each is the right one in a different situation.

Adding and subtracting the addition formulas converts sums into products:

$\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2},\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2},$

with $\cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}$. Turning a sum into a product is almost always the move that makes an equation factorise.

Illustration 1

Find the range of $3\sin \theta +4\cos \theta$.

Write it as $R\sin \left(\theta +\alpha \right)$ with $R=\sqrt{3^2+4^2}=5$ and $\tan \alpha =\frac{4}{3}$. Since the sine ranges over $[-1,1]$, the expression ranges over $[-5,5]$.

In general $a\sin \theta +b\cos \theta$ has range $\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]$, and this single reduction answers every maximum, minimum and range question of that shape.

Illustration 2

Prove that $\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ }=\frac{1}{8}$.

Multiply and divide by $2\sin 20^{\circ }$ and use the double-angle formula repeatedly:

$\frac{2\sin 20^{\circ }\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ }}{2\sin 20^{\circ }}=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{\sin 20^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}.$

The general pattern is $\cos \theta \cos 2\theta \cos 4\theta \cdots =\frac{\sin \left(2^n\theta \right)}{2^n\sin \theta }$, and recognising a chain of doubling angles is what triggers it.

Illustration 3

Find the greatest and least values of ${\sin }^4\theta +{\cos }^4\theta$.

Write it in terms of the double angle. Since ${\sin }^2\theta +{\cos }^2\theta =1$,

${\sin }^4\theta +{\cos }^4\theta ={\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2{\sin }^2\theta {\cos }^2\theta =1-\frac{{\sin }^{2}2\theta }{2}.$

As ${\sin }^{2}2\theta$ runs over $[0,1]$, the expression runs over $\left[\frac{1}{2},1\right]$, attaining $1$ at multiples of $\frac{\pi }{2}$ and $\frac{1}{2}$ midway between them.

The pattern generalises: any symmetric expression in ${\sin }^2\theta$ and ${\cos }^2\theta$ reduces to a function of ${\sin }^{2}2\theta$ alone, after which the range is read off directly.

2. Periodicity and graphs

The sine and cosine repeat every $2\pi$; the tangent and cotangent repeat every $\pi$, because adding $\pi$ negates both the sine and the cosine and their ratio is unchanged. Applying a modulus or squaring halves a period whenever it removes a sign change, so $\left|\sin x\right|$ and ${\sin }^{2}x$ both have period $\pi$ rather than $2\pi$.

Scaling the argument scales the period inversely: $\sin kx$ has period $\frac{2\pi }{\left|k\right|}$. For a sum, the period is a common multiple of the individual periods, and as in the chapter on functions it need not be the smallest one — but for genuinely different frequencies it usually is.

Illustration 4

Find the fundamental period of $\sin \frac{x}{2}+\cos \frac{x}{3}$.

The two periods are $4\pi$ and $6\pi$, whose least common multiple is $12\pi$. Testing $6\pi$: the first term becomes $\sin \left(\frac{x}{2}+3\pi \right)=-\sin \frac{x}{2}$, which changes the sum, so $6\pi$ fails. Testing $4\pi$ likewise fails on the second term.

The period is therefore $12\pi$. Had the two frequencies been incommensurable, as with $\sin x+\sin \sqrt{2}x$, no common multiple would exist and the sum would not be periodic at all.

3. General solutions

A trigonometric equation has infinitely many solutions, and the general forms encode which:

$\sin \theta =\sin \alpha \Rightarrow \theta =n\pi +(-1{)}^n\alpha ,\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha ,$

with $\tan \theta =\tan \alpha \Rightarrow \theta =n\pi +\alpha$. The three differ because the graphs repeat differently: the cosine is even about the origin, the tangent has period $\pi$ rather than $2\pi$, and the sine's alternating sign records its reflection about $\theta =\frac{\pi }{2}$.

Illustration 5

Solve $\sin \theta +\sin 3\theta +\sin 5\theta =0$.

Pair the outer terms and convert to a product: $\sin 5\theta +\sin \theta =2\sin 3\theta \cos 2\theta$. The equation becomes

$\sin 3\theta \left(2\cos 2\theta +1\right)=0.$

So either $\sin 3\theta =0$, giving $\theta =\frac{n\pi }{3}$, or $\cos 2\theta =-\frac{1}{2}$, giving $2\theta =2n\pi \pm \frac{2\pi }{3}$ and hence $\theta =n\pi \pm \frac{\pi }{3}$.

Pairing the terms whose arguments are symmetric about the middle one is what makes the common factor appear; pairing the first two instead leads nowhere.

Illustration 6

Solve $\sqrt{3}\cos \theta +\sin \theta =1$ and count the solutions in $[0,2\pi ]$.

Divide by $R=2$: $\cos \left(\theta -\frac{\pi }{6}\right)=\frac{1}{2}$, so $\theta -\frac{\pi }{6}=2n\pi \pm \frac{\pi }{3}$ and

$\theta =2n\pi +\frac{\pi }{2}\text{or}\theta =2n\pi -\frac{\pi }{6}.$

In $[0,2\pi ]$ these give $\theta =\frac{\pi }{2}$ and $\theta =\frac{11\pi }{6}$: two solutions. Reducing to a single trigonometric function before solving is what keeps the count reliable.

4. Squaring, and the roots it invents

Squaring both sides of an equation is often the only way forward, and it always risks introducing solutions of the negated equation. Every root obtained after squaring must be substituted back into the original.

Illustration 7

Solve $\sin \theta +\cos \theta =1$ for $\theta \in [0,2\pi ]$.

Squaring gives $1+2\sin \theta \cos \theta =1$, so $\sin 2\theta =0$ and $\theta \in \left\{0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi \right\}$.

Testing each in the original: $0$ gives $1$, $\frac{\pi }{2}$ gives $1$, $\pi$ gives $-1$, $\frac{3\pi }{2}$ gives $-1$, and $2\pi$ gives $1$. So the solutions are $0$, $\frac{\pi }{2}$ and $2\pi$; the other two solve $\sin \theta +\cos \theta =-1$ and were manufactured by the squaring.

The alternative route avoids the problem entirely: writing the left side as $\sqrt{2}\sin \left(\theta +\frac{\pi }{4}\right)=1$ produces only genuine solutions.

5. Inverse functions and their principal ranges

Each inverse function is defined by restricting its parent to a stretch on which the parent is one-to-one. The choices are conventional but fixed.

Three complementary relations follow immediately from the ranges: ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ on $[-1,1]$, ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$ on all of $\mathbb{R}$, and ${\sec }^{-1}x+{\mathrm{cosec}}^{-1}x=\frac{\pi }{2}$ for $\left|x\right|\ge 1$. Each pairs a function whose range starts at zero with one centred on zero, which is exactly what makes the two halves fit together into a right angle.

The remaining two inverses have ranges with a hole in them: ${\sec }^{-1}$ takes values in $[0,\pi ]$ excluding $\frac{\pi }{2}$, and ${\mathrm{cosec}}^{-1}$ in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ excluding $0$, because those are precisely the angles at which the parent functions are undefined.

The consequence is that ${\sin }^{-1}\left(\sin x\right)$ equals $x$ only when $x$ already lies in the principal range, and otherwise equals whichever angle in that range has the same sine. The composite is a zigzag, not a straight line.

Illustration 8

Evaluate ${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)$ and ${\sin }^{-1}\left(\sin \frac{5\pi }{6}\right)$.

For the first, $\frac{7\pi }{6}$ is outside $[0,\pi ]$, but $\cos \frac{7\pi }{6}=\cos \frac{5\pi }{6}$ and $\frac{5\pi }{6}$ is inside, so the answer is $\frac{5\pi }{6}$.

For the second, $\frac{5\pi }{6}$ is outside $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, and $\sin \frac{5\pi }{6}=\sin \frac{\pi }{6}$, so the answer is $\frac{\pi }{6}$.

In both cases the method is the same: find the angle inside the principal range with the same value of the direct function.

Illustration 9

Simplify ${\tan }^{-1}\frac{2x}{1-x^2}$ for $|x|<1$ and for $x>1$.

Substituting $x=\tan \theta$ makes the argument $\tan 2\theta$, so the expression is ${\tan }^{-1}\left(\tan 2\theta \right)$.

For $|x|<1$ we have $|\theta |<\frac{\pi }{4}$, so $2\theta$ lies inside the principal range and the answer is $2\theta =2{\tan }^{-1}x$.

For $x>1$ we have $\theta >\frac{\pi }{4}$, so $2\theta >\frac{\pi }{2}$ and lies outside; the answer is $2{\tan }^{-1}x-\pi$. The substitution is standard, but the case analysis afterwards is the part that carries the marks.

Illustration 10

Show that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$ for every $x\in [-1,1]$.

Let $\alpha ={\sin }^{-1}x$, so $\alpha \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and $\sin \alpha =x$. Then $\cos \left(\frac{\pi }{2}-\alpha \right)=\sin \alpha =x$, and $\frac{\pi }{2}-\alpha$ lies in $[0,\pi ]$, which is exactly the principal range of the arccosine. So ${\cos }^{-1}x=\frac{\pi }{2}-\alpha$.

The proof needed the range check at the end. Without it, the conclusion would only say that the two angles have the right cosine, not that one is the principal value.

Two further families of formulas complete the toolkit. The sub-multiple angle relations express everything in terms of $t=\tan \frac{A}{2}$:

$\sin A=\frac{2t}{1+t^2},\cos A=\frac{1-t^2}{1+t^2},\tan A=\frac{2t}{1-t^2},$

which is the same substitution that rationalises trigonometric integrals, and is the quickest route whenever an equation mixes $\sin A$ and $\cos A$ linearly.

The addition formulas for inverse functions carry conditions just as the arctangent one does. For instance

${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)$

holds only when $x^2+y^2\le 1$, or when $xy<0$; outside that region the true sum leaves $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ and the right side must be replaced by $\pi$ minus it, or by $-\pi$ minus it, according to the signs. As always, the algebra is the easy half and the range check is the examined half.

6. Solving triangles

In a triangle with sides $a,b,c$ opposite angles $A,B,C$, and circumradius $R$,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R,\cos A=\frac{b^2+c^2-a^2}{2bc}.$

The sine rule is used when a side and its opposite angle are known; the cosine rule when they are not. The cosine rule is also the safer of the two for finding an angle, because the sine rule leaves an ambiguity between an angle and its supplement, which have the same sine.

The area has several forms, each convenient for different data:

$\Delta =\frac{1}{2}ab\sin C=\sqrt{s(s-a)(s-b)(s-c)}=rs=\frac{abc}{4R},$

where $s$ is the semi-perimeter and $r$ the inradius.

Illustration 11

In a triangle, $a=13$, $b=14$ and $c=15$. Find its area, circumradius and inradius.

The semi-perimeter is $21$, so by Heron's formula

$\Delta =\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84.$

Then $R=\frac{abc}{4\Delta }=\frac{2730}{336}=\frac{65}{8}$ and $r=\frac{\Delta }{s}=\frac{84}{21}=4$.

As a check, $\cos A=\frac{196+225-169}{2\cdot 14\cdot 15}=\frac{252}{420}=\frac{3}{5}$, so $\sin A=\frac{4}{5}$ and $\frac{a}{\sin A}=\frac{13}{4/5}=\frac{65}{4}=2R$, which agrees.

Illustration 12

In a triangle, $b=\sqrt{3}$, $c=1$ and $B=60^{\circ }$. Find $C$ and hence $A$.

By the sine rule, $\sin C=\frac{c\sin B}{b}=\frac{1\cdot \frac{\sqrt{3}}{2}}{\sqrt{3}}=\frac{1}{2}$, so $C=30^{\circ }$ or $150^{\circ }$.

The second is impossible, since $B+C$ would exceed $180^{\circ }$. So $C=30^{\circ }$ and $A=90^{\circ }$. Discarding the supplementary value requires a reason, and here the angle sum supplies it; in other configurations the reason is that the larger side must face the larger angle.

Illustration 13

Prove that in any triangle $a\cos A+b\cos B+c\cos C=\frac{2\Delta }{R}$.

Substitute $a=2R\sin A$ and the corresponding forms for $b$ and $c$:

$a\cos A+b\cos B+c\cos C=2R\left(\sin A\cos A+\sin B\cos B+\sin C\cos C\right)=R\left(\sin 2A+\sin 2B+\sin 2C\right).$

Because $A+B+C=\pi$, the standard conditional identity $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$ applies, so the expression equals $4R\sin A\sin B\sin C$.

Finally $\Delta =\frac{1}{2}ab\sin C=2R^2\sin A\sin B\sin C$, so $\sin A\sin B\sin C=\frac{\Delta }{2R^2}$ and the expression is $\frac{2\Delta }{R}$.

Checking on an equilateral triangle of side $a$: the left side is $\frac{3a}{2}$, while $R=\frac{a}{\sqrt{3}}$ and $\Delta =\frac{\sqrt{3}{a}^2}{4}$ give $\frac{2\Delta }{R}=\frac{3a}{2}$ as well.

Converting every side into $2R\sin (\text{angle})$ turns a relation between sides into a trigonometric identity in the three angles, which is the standard technique for triangle identities.

One relation is worth stating separately because it is so easily forgotten. The projection formula says

$a=b\cos C+c\cos B,$

with two companions obtained by permuting the letters, and it drops out of the sine rule in one line: $a=2R\sin A=2R\sin (B+C)=2R\left(\sin B\cos C+\cos B\sin C\right)=b\cos C+c\cos B$. Geometrically it says that the foot of the altitude from $A$ splits $a$ into the two projections of the other sides.

Illustration 14

In a triangle, prove that ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C+2\cos A\cos B\cos C=1$.

Since $C=\pi -(A+B)$, write $\cos C=-\cos (A+B)$ and treat the whole expression as a quadratic in $\cos C$:

${\cos }^{2}C+2\cos A\cos B\cos C+\left({\cos }^{2}A+{\cos }^{2}B-1\right).$

The constant term is ${\cos }^{2}A-{\sin }^{2}B$, which factorises as $\left(\cos A+\sin B\right)\left(\cos A-\sin B\right)$, and substituting $\cos C=-\cos A\cos B+\sin A\sin B$ makes every term cancel.

Every conditional identity in a triangle is handled this way: eliminate one angle using $A+B+C=\pi$, then the statement becomes an ordinary identity in two free angles.

Illustration 15

In a triangle, prove that $\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$, and use it when $a=5$, $b=6$, $c=7$.

The half-angle formula follows from $\cos A=1-2{\sin }^2\frac{A}{2}$ combined with the cosine rule, which gives ${\sin }^2\frac{A}{2}=\frac{(s-b)(s-c)}{bc}$ and ${\cos }^2\frac{A}{2}=\frac{s(s-a)}{bc}$; dividing gives the stated result.

With $s=9$, $\tan \frac{A}{2}=\sqrt{\frac{3\cdot 2}{9\cdot 4}}=\sqrt{\frac{1}{6}}$. Then $\tan A=\frac{2t}{1-t^2}$ with $t^2=\frac{1}{6}$ gives $\tan A=\frac{2/\sqrt{6}}{5/6}=\frac{12}{5\sqrt{6}}$, and squaring back reproduces $\cos A=\frac{36+49-25}{84}=\frac{5}{7}$, which is consistent.

Summary

The inverse trigonometric functions undo their parents only on one chosen stretch, so ${\sin }^{-1}\left(\sin x\right)$ is a zigzag rather than the identity, and the arctangent addition formula needs a correction of $\pm \pi$ whenever $xy>1$. Every composite question is answered by finding the angle inside the principal range that has the same value of the direct function.

The Pythagorean relations and the two addition formulas generate everything else, including the three forms of $\cos 2A$ and the sum-to-product conversions. Turning a sum into a product is the move that makes an equation factorise, and reducing $a\sin \theta +b\cos \theta$ to a single sine gives both its range and a clean route to its zeros.

Periods halve when a modulus or a square removes a sign change, and the period of a sum is a common multiple of the individual periods rather than automatically the smallest one.

General solutions differ between the three functions because the graphs repeat differently: $n\pi +(-1{)}^n\alpha$ for the sine, $2n\pi \pm \alpha$ for the cosine, and $n\pi +\alpha$ for the tangent. Squaring an equation introduces the solutions of its negation, so every root found that way must be substituted back.

In a triangle, the sine rule relates a side to its opposite angle and to $2R$, while the cosine rule is safer for finding an angle because it distinguishes an angle from its supplement. The area has four standard forms, and converting each side into $2R\sin (\text{angle})$ turns any relation between sides into an identity in the three angles.