Trigonometry

Simplify ${\sin }^{-1}(\sin 2)$, with $2$ in radians.

The functions undo each other, so the answer is $2$. That is what everybody writes.

$\sin 2=0.9093,{\sin }^{-1}(0.9093)=1.1416$

The answer is $\pi -2$, not $2$.

Nothing went wrong with the arithmetic. Two radians is about $114.6^{\circ }$, and ${\sin }^{-1}$ is only permitted to return values between $-\pi /2$ and $\pi /2$. It cannot return $2$, because $2$ is outside its range, so it returns the angle in range with the same sine.

The same feature causes the other half of the chapter's difficulty. Solve $\sin \theta =\frac{1}{2}$ and the answer is not $\pi /6$; it is $\pi /6$ and $5\pi /6$ and $13\pi /6$ and infinitely many more.

$\theta =n\pi +(-1{)}^n\frac{\pi }{6},n\in \mathbb{Z}$

The trigonometric functions are periodic, so they are many-to-one. Everything difficult in this chapter is bookkeeping about that one fact.

1. Angles, Radians and the Ratios

$\pi \text{ radians}=180^{\circ },\text{arc }\ell =r\theta ,\text{sector area}=\frac{1}{2}{r}^2\theta$

Those last two hold only in radians, which is one reason radians are the default everywhere in calculus.

Signs matter more than values in this chapter, because most errors are a correct magnitude with the wrong sign attached.

2. The Fundamental Identities

${\sin }^2\theta +{\cos }^2\theta =1,1+{\tan }^2\theta ={\sec }^2\theta ,1+{\cot }^2\theta ={\csc }^2\theta$

The second and third are the first divided by ${\cos }^2\theta$ and ${\sin }^2\theta$, so there is really one identity here, not three.

$\sin (-\theta )=-\sin \theta ,\cos (-\theta )=\cos \theta ,\tan (-\theta )=-\tan \theta$

Cosine is even; sine and tangent are odd. That single line settles most sign questions faster than the quadrant table.

3. Compound, Multiple and Sub-Multiple Angles

$\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B$

$\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$

Trap. The signs are reversed in the cosine formula: $\cos (A+B)$ takes a minus. Checking with $A=B=0$ will not catch it; check with $A=B=\pi /2$, where $\cos \pi =-1$ and only the correct signs deliver it.

$\sin 2\theta =2\sin \theta \cos \theta ,\cos 2\theta =1-2{\sin }^2\theta =2{\cos }^2\theta -1$

$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta },\sin 3\theta =3\sin \theta -4{\sin }^3\theta ,\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

Illustration 1

Find the exact value of $\sin 15^{\circ }$, and derive $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$.

Write $15^{\circ }$ as a difference of angles you already know.

$\sin 15^{\circ }=\sin (45^{\circ }-30^{\circ })=\sin 45\cos 30-\cos 45\sin 30=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot \frac{1}{2}$

$=\frac{\sqrt{6}-\sqrt{2}}{4}\approx \frac{2.449-1.414}{4}=0.2588$

For the triple angle, split $3\theta$ as $2\theta +\theta$ and expand each piece.

$\sin 3\theta =\sin 2\theta \cos \theta +\cos 2\theta \sin \theta =2\sin \theta {\cos }^2\theta +(1-2{\sin }^2\theta )\sin \theta$

Replace ${\cos }^2\theta$ by $1-{\sin }^2\theta$ so that only sines remain, which is the point of the formula.

$=2\sin \theta -2{\sin }^3\theta +\sin \theta -2{\sin }^3\theta =3\sin \theta -4{\sin }^3\theta$

Every multiple-angle formula is built this way, so none of them needs to be memorised separately.

Illustration 2

If $A+B+C=\pi$, prove that $\tan A+\tan B+\tan C=\tan A\tan B\tan C$.

Use the condition to express one angle in terms of the others, then take tangents.

$A+B=\pi -C\Rightarrow \tan (A+B)=\tan (\pi -C)=-\tan C$

Now expand the left side by the compound-angle formula.

$\frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$

Cross-multiply and collect the terms, which is the whole of the remaining work.

$\tan A+\tan B=-\tan C+\tan A\tan B\tan C$

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

The identity looks striking but the proof used only one fact: that the tangent of a supplement is the negative of the tangent. Every conditional identity of this kind works the same way, by converting the constraint into a statement about one ratio and then expanding.

4. Transformations: Sums and Products

$\cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2},\cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}$

$\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2},\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$

A sum cannot be factorised; a product can. That is the whole reason to convert: an equation like $\cos 3\theta +\cos \theta =0$ is unsolvable as written and trivial once it becomes a product.

Illustration 3

Solve $\cos 3\theta +\cos \theta =0$.

Expanding $\cos 3\theta$ into cubes of $\cos \theta$ would give a cubic. Convert the sum to a product instead.

$2\cos \frac{3\theta +\theta }{2}\cos \frac{3\theta -\theta }{2}=2\cos 2\theta \cos \theta =0$

A product is zero when either factor is, so the equation splits cleanly into two.

$\cos 2\theta =0\Rightarrow 2\theta =(2n+1)\frac{\pi }{2}\Rightarrow \theta =(2n+1)\frac{\pi }{4}$

$\cos \theta =0\Rightarrow \theta =(2n+1)\frac{\pi }{2}$

Both families are needed; neither contains the other, since $\pi /4$ appears only in the first and $\pi /2$ only in the second.

5. The Range of $a\sin \theta +b\cos \theta$

$a\sin \theta +b\cos \theta =R\sin (\theta +\varphi ),R=\sqrt{a^2+b^2}$

Two waves of the same frequency always add to a single wave of that frequency, with a new amplitude and a phase shift. Since a sine never leaves $[-1,1]$:

$-\sqrt{a^2+b^2}\le a\sin \theta +b\cos \theta \le \sqrt{a^2+b^2}$

Trap. The maximum is $\sqrt{a^2+b^2}$, not $a+b$. The two terms peak at different values of $\theta$, so they can never both be at their maximum together.

Illustration 4

Find the maximum and minimum of $3\sin \theta +4\cos \theta +5$.

The variable part has amplitude given by the square root of the sum of the squares.

$R=\sqrt{9+16}=5$

$-5\le 3\sin \theta +4\cos \theta \le 5$

Adding the constant $5$ shifts the whole range without changing its width.

$0\le 3\sin \theta +4\cos \theta +5\le 10$

The maximum is $10$ and the minimum is $0$. Note that the naive $3+4+5=12$ would have been wrong by two, and that the minimum landing exactly on zero is a coincidence of these numbers rather than a general feature.

6. Trigonometric Equations and General Solutions

Each formula is the shape of that function's graph written in symbols. Sine is symmetric about $\pi /2$, so its solutions alternate; cosine is symmetric about the $y$-axis, so they come in $\pm$ pairs; tangent has period $\pi$ rather than $2\pi$, so its solutions are evenly spaced.

The method: reduce the equation to a single ratio of a single angle, then apply the matching row.

Illustration 5

Solve $2{\cos }^2\theta +3\sin \theta =0$.

Two different ratios appear, so convert one into the other before anything else. The Pythagorean identity turns cosine squared into sine squared.

$2(1-{\sin }^2\theta )+3\sin \theta =0\Rightarrow 2{\sin }^2\theta -3\sin \theta -2=0$

$(2\sin \theta +1)(\sin \theta -2)=0$

$\sin \theta =-\frac{1}{2}\text{or}\sin \theta =2$

Reject $\sin \theta =2$. A sine never exceeds $1$, so that factor contributes nothing, and stating the rejection is part of a complete answer.

$\theta =n\pi +(-1{)}^n\left(-\frac{\pi }{6}\right)$

Illustration 6

Solve $\sin \theta +\cos \theta =1$, and explain why squaring must be checked.

Squaring both sides removes the mixed ratios, and it is the obvious move.

$1+2\sin \theta \cos \theta =1\Rightarrow \sin 2\theta =0\Rightarrow 2\theta =n\pi \Rightarrow \theta =\frac{n\pi }{2}$

That offers $0$, $\pi /2$, $\pi$ and $3\pi /2$ within one revolution. Test each in the original equation.

Half of them are false. Squaring turns $X=1$ into $X^2=1$, which also admits $X=-1$, so it manufactures solutions to a different equation.

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

Trap. Any step that squares, or multiplies by something that can vanish, can create false roots. Substitute every candidate back into the original.

Illustration 7

Solve $\sin 2\theta =\cos 3\theta$.

The two sides use different ratios, so no general-solution row applies yet. Convert one into the other using the complementary relation.

$\cos 3\theta =\sin \left(\frac{\pi }{2}-3\theta \right)\Rightarrow \sin 2\theta =\sin \left(\frac{\pi }{2}-3\theta \right)$

Now the sine row applies, with $\alpha =\frac{\pi }{2}-3\theta$.

$2\theta =n\pi +(-1{)}^n\left(\frac{\pi }{2}-3\theta \right)$

The alternating sign means the even and odd cases must be handled separately.

$n=2k:2\theta =2k\pi +\frac{\pi }{2}-3\theta \Rightarrow 5\theta =2k\pi +\frac{\pi }{2}\Rightarrow \theta =\frac{2k\pi }{5}+\frac{\pi }{10}$

$n=2k+1:2\theta =(2k+1)\pi -\frac{\pi }{2}+3\theta \Rightarrow \theta =\frac{\pi }{2}-(2k+1)\pi$

Both families are part of the answer. Splitting on the parity of $n$ is compulsory whenever the sine row is used with the unknown appearing on both sides, because the two cases give genuinely different equations.

7. Inverse Trigonometric Functions: The Branch Problem

A periodic function is many-to-one, so it has no inverse at all until its domain is cut down to a stretch on which it is one-to-one. The chosen stretch is the principal branch.

${\sin }^{-1}(\sin x)=x\text{only if }x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Outside that, reduce first: use $\sin (\pi -x)=\sin x$ to bring the angle into range, and similarly $\cos (2\pi -x)=\cos x$ and $\tan (x-\pi )=\tan x$.

Illustration 8

Evaluate ${\sin }^{-1}(\sin 2)$, ${\cos }^{-1}(\cos (-1))$ and ${\tan }^{-1}(\tan 3)$, all in radians.

Check each argument against the relevant range before doing anything else.

${\sin }^{-1}(\sin 2)$. Since $2>\pi /2\approx 1.571$, the answer cannot be $2$. Use $\sin 2=\sin (\pi -2)$, and $\pi -2\approx 1.142$ does lie in $[-\pi /2,\pi /2]$.

${\sin }^{-1}(\sin 2)=\pi -2$

${\cos }^{-1}(\cos (-1))$. Since $-1<0$, it is outside $[0,\pi ]$. Cosine is even, so $\cos (-1)=\cos 1$, and $1$ is in range.

${\cos }^{-1}(\cos (-1))=1$

${\tan }^{-1}(\tan 3)$. Since $3>\pi /2$, subtract the period $\pi$: $3-\pi \approx -0.142$, comfortably inside $(-\pi /2,\pi /2)$.

${\tan }^{-1}(\tan 3)=3-\pi$

Three different adjustments, because the three ranges are different. Always compare the argument with the range before writing anything.

8. Properties of the Inverse Functions

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2},{\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2},{\sec }^{-1}x+{\csc }^{-1}x=\frac{\pi }{2}$

${\sin }^{-1}(-x)=-{\sin }^{-1}x,{\cos }^{-1}(-x)=\pi -{\cos }^{-1}x$

Note the asymmetry in that last pair: the odd functions simply flip sign, while ${\cos }^{-1}$ reflects about $\pi /2$ because its range is $[0,\pi ]$ rather than a symmetric interval.

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}\text{only if }xy<1$

Trap. That condition is not decoration. When $xy>1$ the true answer differs from the formula by $\pi$, because the sum has left the principal range.

Illustration 9

Evaluate ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$, and then ${\tan }^{-1}2+{\tan }^{-1}3$.

For the first, check the condition: $xy=\frac{1}{6}<1$, so the formula applies directly.

${\tan }^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}={\tan }^{-1}\frac{5/6}{5/6}={\tan }^{-1}1=\frac{\pi }{4}$

Numerically, $0.4636+0.3217=0.7854=\pi /4$, confirming it.

For the second, $xy=6>1$, so the formula alone is wrong.

$\frac{x+y}{1-xy}=\frac{5}{1-6}=-1\Rightarrow {\tan }^{-1}(-1)=-\frac{\pi }{4}$

But both original terms are positive and each exceeds $\pi /4$, so their sum must exceed $\pi /2$ and certainly cannot be negative. Add $\pi$ to bring it back.

${\tan }^{-1}2+{\tan }^{-1}3=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

Check: $1.1071+1.2490=2.3562=3\pi /4$. Estimate the size of the answer before trusting the formula, and the correction becomes obvious rather than arbitrary.

Illustration 10

If ${\sin }^{-1}x+{\sin }^{-1}y=\frac{\pi }{2}$, show that $x^2+y^2=1$.

Rearrange so that one inverse sine stands alone, then use the complementary identity.

${\sin }^{-1}y=\frac{\pi }{2}-{\sin }^{-1}x={\cos }^{-1}x$

Take the sine of both sides. The right needs $\sin ({\cos }^{-1}x)$, which is the sine of an angle whose cosine is $x$.

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

$y^2=1-x^2\Rightarrow x^2+y^2=1$

The positive root is correct because ${\cos }^{-1}x$ lies in $[0,\pi ]$, where the sine is never negative. That range check is what makes the step legitimate rather than a guess between two signs.

Illustration 11

Show that ${\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\pi$.

The previous illustration already established the awkward pair, so use it rather than starting again.

${\tan }^{-1}2+{\tan }^{-1}3=\frac{3\pi }{4}$

${\tan }^{-1}1+\frac{3\pi }{4}=\frac{\pi }{4}+\frac{3\pi }{4}=\pi$

Numerically: $0.7854+1.1071+1.2490=3.1416$.

The result is a small surprise worth sitting with. Each term is a perfectly ordinary angle under $\pi /2$, yet the three sum to a straight angle exactly, with no approximation anywhere. Combining them in the other order works too, provided each application of the addition formula is checked against the $xy<1$ condition: ${\tan }^{-1}1+{\tan }^{-1}2$ has $xy=2>1$ and therefore needs its own correction by $\pi$.

9. A Note on Syllabus Emphasis

The unit text names trigonometric identities and equations, trigonometric functions, and inverse trigonometric functions with their properties.

Heights and distances and the properties of triangles appear throughout older books and question banks, and remain examinable in JEE Advanced. For Main, the marks sit in general solutions and in the inverse functions, and within the inverse functions they sit almost entirely on the range restrictions.

Summary

The trigonometric functions are periodic, so they are many-to-one, and every difficulty here is bookkeeping about that.

An inverse function exists only after the domain is cut to a principal branch, so ${\sin }^{-1}(\sin x)=x$ only for $x$ in $[-\pi /2,\pi /2]$. Outside it, reduce the angle into range first.

The three ranges differ, so the three adjustments differ: $\pi -x$ for sine, evenness for cosine, subtracting $\pi$ for tangent.

Arc length and sector area hold only in radians. Cosine is even, sine and tangent are odd, and that settles most sign questions.

There is really one Pythagorean identity; the other two are it divided through.

The compound-angle cosine formula reverses the signs, and every multiple-angle formula is built from the compound ones rather than memorised.

Convert sums to products when solving, because a sum cannot be factorised and a product splits into cases immediately.

$a\sin \theta +b\cos \theta$ has amplitude $\sqrt{a^2+b^2}$, never $a+b$, because the two terms peak at different angles.

General solutions: sine alternates, cosine comes in $\pm$ pairs, tangent has period $\pi$. Reduce to one ratio of one angle first.

Reject impossible roots such as $\sin \theta =2$, and say that you have.

Squaring manufactures solutions to a different equation, so substitute every candidate back into the original: $\sin \theta +\cos \theta =1$ loses half of them.

Conditional identities are proved by turning the constraint into a statement about one ratio, then expanding.

When the unknown appears on both sides of a sine equation, split on the parity of $n$ before solving.

${\tan }^{-1}x+{\tan }^{-1}y$ needs $xy<1$; beyond that the answer differs by $\pi$. Estimate the size of the answer before trusting any inverse formula.