Trigonometric Functions

The chapter opens by fixing units of rotation. Section 3.2 defines an angle as a measure of rotation, introduces degree and radian measure, and proves $\pi$ radian $= 180°$ along with the arc-length formula $l = r\theta$ — the machinery behind Examples 1 to 5 on converting angles and finding arc lengths and radii. Section 3.3 is the conceptual heart: it re-defines the trigonometric functions using a point moving on a unit circle, works out exact values at the standard and quadrantal angles, and tabulates sign, domain, range and increasing/decreasing behaviour — the ground for Examples 6 to 9, finding the remaining five functions from one given ratio and a stated quadrant. Section 3.4 is where the chapter earns its length: trigonometric functions of the sum and difference of two angles. It proves the compound-angle formula for $\cos(x+y)$ from first principles, then derives, by substitution alone, a cascade of about twenty numbered identities — difference formulas, allied-angle results, $\tan(x\pm y)$ and $\cot(x\pm y)$, the double- and triple-angle formulas, and the sum-to-product and product-to-sum transformations. NCERT's Examples 10 to 22 and the Miscellaneous Examples are almost entirely applications of this section — evaluating exact values such as $\sin 15°$, proving identities, and finding half-angle ratios. This article follows the same arc: angle measure, the unit-circle definition, then the full compound-angle machinery.

Angle measure: degrees, radians and arc length

An angle is a measure of rotation of a ray about its endpoint, from an initial side to a terminal side, with the fixed point called the vertex. Rotation counted anticlockwise gives a positive angle; rotation counted clockwise gives a negative angle. Degree measure divides one full revolution into $360$ equal parts, each called a degree ($1°$), further split into $60$ minutes ($1° = 60'$) and each minute into $60$ seconds ($1' = 60''$). Radian measure instead ties the angle directly to a circle: the angle subtended at the centre of a circle by an arc equal in length to the radius is defined as one radian. Because equal arcs of a circle subtend equal angles at the centre, an arc of length $l$ in a circle of radius $r$ subtends an angle $\theta$ (in radians) given by

where: $l$ is the arc length, $r$ is the radius, and $\theta$ is measured in radians, not degrees — this formula is only valid with $\theta$ in radians, since it comes directly from the definition of one radian as "arc length equal to radius." Since the full circumference of a unit circle is $2\pi$, one complete revolution subtends an angle of exactly $2\pi$ radians, and this single fact fixes the conversion between the two units: $2\pi$ radian $= 360°$, so

where: the two conversion rules in (2) are inverses of each other, and by convention an unlabelled numerical angle (such as "$\pi/4$") is always understood to be in radians, while an angle written with the degree symbol is in degrees. A short table of the most-used conversions, reused constantly through the rest of the chapter, is worth memorising outright:

The unit circle definition of the trigonometric functions

Class 9 and 10 defined sine and cosine only as ratios of sides in a right triangle, which restricts the angle to strictly between $0°$ and $90°$. This chapter replaces that definition with one that works for every real number. Take the unit circle (radius $1$) centred at the origin, and let $P(a,b)$ be the point reached by moving an arc of length $x$ anticlockwise from $A(1,0)$ — so $x$ may be any real number, positive, negative, or larger than $2\pi$. The two basic functions are defined as the coordinates of this point:

where: $a$ and $b$ are the horizontal and vertical coordinates of $P$, and $x$ is the arc length (equivalently, the radian measure of $\angle AOP$) that located $P$. Since $\triangle OMP$ (with $M$ the foot of the perpendicular from $P$ to the x-axis) is right-angled, Pythagoras gives $OM^2+MP^2=OP^2$, i.e. $a^2+b^2=1$, which is exactly

where: equation (4) is the chapter's single most-used identity, valid unconditionally — there is no restriction on $x$, unlike the right-triangle definition it replaces. The remaining four functions are defined in terms of these two:

where: dividing identity (4) through by $\cos^2 x$ or by $\sin^2 x$ produces the two companion Pythagorean identities used constantly alongside it, $1+\tan^2x=\sec^2x$ and $1+\cot^2x=\text{cosec}^2x$. Because $A, B, C, D$ at angles $0, \pi/2, \pi, 3\pi/2$ (the quadrantal angles, integral multiples of $\pi/2$) have coordinates $(1,0),(0,1),(-1,0),(0,-1)$, the values of sine and cosine there follow immediately from the definition without any triangle at all: $\sin 0 = 0$, $\cos 0 = 1$, $\sin\tfrac{\pi}{2}=1$, $\cos\tfrac{\pi}{2}=0$, $\sin\pi=0$, $\cos\pi=-1$, and so on. Combined with the familiar right-triangle values at $30°, 45°, 60°$, this gives the complete standard-angle table used throughout the chapter:

Domain, range, sign and periodicity

Since every real number $x$ locates a point on the unit circle, $\sin x$ and $\cos x$ are defined for all real $x$, and since $-1\le a\le1$ and $-1\le b\le1$ for any point on the unit circle, their range is exactly $[-1,1]$. The four derived functions inherit domain restrictions from their denominators: $\tan x$ and $\sec x$ are undefined wherever $\cos x = 0$, that is at $x = (2n+1)\tfrac{\pi}{2}$ for integer $n$; and $\cot x$ and $\text{cosec}\ x$ are undefined wherever $\sin x = 0$, that is at $x = n\pi$ for integer $n$. The sign of each function in a given quadrant follows directly from the signs of $a$ and $b$ there, and is summarised by the mnemonic All-Sine-Tangent-Cosine (ASTC): all six functions are positive in quadrant I; only sine and cosec are positive in quadrant II; only tangent and cotangent are positive in quadrant III; only cosine and secant are positive in quadrant IV.

Because retracing a full revolution returns $P$ to the same point, $\sin(2n\pi + x) = \sin x$ and $\cos(2n\pi + x) = \cos x$ for every integer $n$ — sine and cosine are periodic functions with period $2\pi$ (the smallest positive value that repeats the function). Since $\text{cosec}\ x$ and $\sec x$ are their reciprocals, they too have period $2\pi$. Tangent and cotangent turn out to have the shorter period $\pi$ — a fact proved in the compound-angle section below, since $\tan(\pi+x)=\tan x$.

Compound-angle formulas and the allied angles

The chapter's real engine is a single geometric proof: the formula for $\cos(x+y)$ in terms of $\cos x, \sin x, \cos y, \sin y$. It is proved once (in the first derivation below) and then, by pure substitution — replacing $y$ by $-y$, swapping the roles of $x$ and $y$, or choosing special values — it generates roughly twenty further identities without any new geometry. The two headline results are

where: these four hold for all real $x$ and $y$, with no restriction — the sine formulas in (7) are themselves derived from the cosine formulas in (6) using the allied-angle results $\cos\left(\tfrac{\pi}{2}-x\right)=\sin x$ and $\sin\left(\tfrac{\pi}{2}-x\right)=\cos x$. Dividing the sine formula by the cosine formula, then dividing numerator and denominator by $\cos x\cos y$, gives the tangent versions, valid whenever none of $x$, $y$, $x\pm y$ is an odd multiple of $\pi/2$:

where: the numerator sign always matches the sign inside $\tan(x\pm y)$, while the denominator sign is always the opposite — a common exam slip. A parallel pair of formulas exists for $\cot(x\pm y)$, obtained the same way from (6) divided by (7). Choosing special values of $x$ and $y$ in (6)–(8) produces the full family of allied-angle identities relating a function of $\left(\tfrac{\pi}{2}\pm x\right)$, $(\pi\pm x)$ or $(2\pi - x)$ back to a function of $x$ alone — for instance $\cos(\pi+x)=-\cos x$, $\sin(\pi-x)=\sin x$, and $\tan(\pi+x)=\tan x$, the last of which proves tangent's period is $\pi$, not $2\pi$.

Multiple and sub-multiple angle formulas

Setting $y=x$ inside the compound-angle formulas (6)–(8) collapses "angle of a sum" into "angle doubled," giving the double-angle formulas. Because $\cos^2x+\sin^2x=1$, the cosine version can be rewritten in three interchangeable forms, all equally standard:

where: the three forms of $\cos 2x$ are not three different facts to memorise separately but one identity written three ways using $\cos^2x+\sin^2x=1$ to swap between $\sin^2x$ and $\cos^2x$ — choose whichever form matches the information given in a problem (a value of $\sin x$ alone, $\cos x$ alone, or $\tan x$ alone). Writing $3x = 2x+x$ and substituting the double-angle results into the compound-angle formulas (a full derivation is given below) produces the triple-angle formulas:

where: these are used constantly in JEE-level identity proofs (for example $\tan 3x\tan 2x\tan x = \tan 3x-\tan 2x-\tan x$, NCERT's own Example 14) and, run in reverse with $x$ replaced by $x/2$, they give the half-angle formulas $\sin\tfrac{x}{2}=\pm\sqrt{\tfrac{1-\cos x}{2}}$ and $\cos\tfrac{x}{2}=\pm\sqrt{\tfrac{1+\cos x}{2}}$ — genuinely ambiguous in sign, resolved only by knowing which quadrant $x/2$ lies in, exactly the subtlety tested in the competitive-level worked example below.

Sum-to-product and product-to-sum formulas

Adding and subtracting the four compound-angle formulas (6)–(7) in pairs converts a product of two cosines or two sines into a sum, the product-to-sum formulas:

where: these four are exactly identities (6) and (7) added or subtracted in pairs, nothing more. Substituting $x+y=\theta$ and $x-y=\varphi$ (so $x=\tfrac{\theta+\varphi}{2}$, $y=\tfrac{\theta-\varphi}{2}$) and relabelling turns (11)–(12) inside out into the sum-to-product formulas, which convert a sum or difference of two sines or cosines into a product — the full derivation is given below:

where: these are the workhorse tools for simplifying expressions like $\dfrac{\sin 5x+\sin x}{\cos 5x+\cos x}$ into $\tan 3x$ (NCERT's Example 19 pattern) — the rule of thumb is: whenever a fraction has sums or differences of sines/cosines in both numerator and denominator, apply (13)–(14) first, and the common factor $2\cos\tfrac{x+y}{2}$ or $2\sin\tfrac{x+y}{2}$ usually cancels between top and bottom.

Board level - finding all six trigonometric functions from one ratio and a quadrant (extends NCERT Example 7)

If $\cot x = -\dfrac{5}{12}$ and $x$ lies in the second quadrant, find the values of the other five trigonometric functions.

Step 1. Since $\tan x$ is the reciprocal of $\cot x$, $\tan x = -\dfrac{12}{5}$. This matches the ASTC rule: in the second quadrant only sine and cosec are positive, so tangent must indeed come out negative, a useful sanity check before proceeding.

Step 2. Use $1+\tan^2x=\sec^2x$: $\sec^2x = 1+\dfrac{144}{25} = \dfrac{169}{25}$, so $\sec x = \pm\dfrac{13}{5}$. In the second quadrant cosine (and hence secant) is negative, so $\sec x = -\dfrac{13}{5}$, giving $\cos x = -\dfrac{5}{13}$.

Step 3. Recover $\sin x$ from $\sin x = \tan x\cos x = \left(-\dfrac{12}{5}\right)\times\left(-\dfrac{5}{13}\right) = \dfrac{12}{13}$, positive as expected in the second quadrant, and take its reciprocal for $\text{cosec}\ x = \dfrac{13}{12}$.

Step 4. Check with the Pythagorean identity: $\cos^2x+\sin^2x = \dfrac{25}{169}+\dfrac{144}{169} = \dfrac{169}{169}=1$. Answer: $\sin x = \mathbf{\dfrac{12}{13}}$, $\cos x = \mathbf{-\dfrac{5}{13}}$, $\tan x = \mathbf{-\dfrac{12}{5}}$, $\text{cosec}\ x = \mathbf{\dfrac{13}{12}}$, $\sec x = \mathbf{-\dfrac{13}{5}}$.

Board level - exact values of sin 15°, cos 15° and tan 15° using compound-angle formulas (extends NCERT Examples 11 and 12)

Using the compound-angle formulas and the standard angles $45°$ and $30°$, find the exact values of $\sin 15°$, $\cos 15°$ and $\tan 15°$.

Step 1. Write $15° = 45°-30°$ and apply the sine difference formula: $\sin 15° = \sin(45°-30°) = \sin 45°\cos 30°-\cos 45°\sin 30° = \dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2}\cdot\dfrac{1}{2} = \dfrac{\sqrt3-1}{2\sqrt2} = \dfrac{\sqrt6-\sqrt2}{4}$.

Step 2. Apply the cosine difference formula the same way: $\cos 15° = \cos(45°-30°) = \cos 45°\cos 30°+\sin 45°\sin 30° = \dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}+\dfrac{1}{\sqrt2}\cdot\dfrac{1}{2} = \dfrac{\sqrt3+1}{2\sqrt2} = \dfrac{\sqrt6+\sqrt2}{4}$.

Step 3. Find $\tan 15°$ either as $\dfrac{\sin 15°}{\cos 15°}$ or directly from the tangent difference formula, which is quicker: $\tan 15° = \tan(45°-30°) = \dfrac{\tan45°-\tan30°}{1+\tan45°\tan30°} = \dfrac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}} = \dfrac{\sqrt3-1}{\sqrt3+1}$.

Step 4. Rationalise by multiplying numerator and denominator by $(\sqrt3-1)$: $\tan 15° = \dfrac{(\sqrt3-1)^2}{(\sqrt3)^2-1^2} = \dfrac{3-2\sqrt3+1}{2} = \dfrac{4-2\sqrt3}{2} = 2-\sqrt3$. Answer: $\sin 15° = \mathbf{\dfrac{\sqrt6-\sqrt2}{4}}$, $\cos 15° = \mathbf{\dfrac{\sqrt6+\sqrt2}{4}}$, $\tan 15° = \mathbf{2-\sqrt3}$ — the same numerical value NCERT reaches independently for $\tan\dfrac{13\pi}{12}$, since $\dfrac{13\pi}{12}$ and $15°=\dfrac{\pi}{12}$ give the same tangent by the period-$\pi$ property.

Competitive level (JEE / NEET) - half-angle ratios with the quadrant sign trap (extends NCERT Example 21)

If $\tan x = \dfrac{3}{4}$ and $\pi < x < \dfrac{3\pi}{2}$, find $\sin\dfrac{x}{2}$, $\cos\dfrac{x}{2}$ and $\tan\dfrac{x}{2}$.

Step 1. Locate $x/2$ first, before touching any formula — this is the step most students skip and the one that decides every sign in the answer. Since $\pi < x < \dfrac{3\pi}{2}$, dividing throughout by $2$ gives $\dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3\pi}{4}$, which places $x/2$ in the second quadrant. So $\sin(x/2)$ must come out positive, while $\cos(x/2)$ and $\tan(x/2)$ must come out negative — fix this now, before computing anything, as a check against the algebra below.

Step 2. Find $\cos x$ from $\tan x$: $\sec^2x = 1+\tan^2x = 1+\dfrac{9}{16} = \dfrac{25}{16}$, so $\sec x=\pm\dfrac{5}{4}$. Since $x$ itself lies in the third quadrant ($\pi < x < 3\pi/2$), cosine is negative there, so $\sec x = -\dfrac{5}{4}$ and $\cos x = -\dfrac{4}{5}$.

Step 3. Use the double-angle identities in reverse: $2\sin^2\dfrac{x}{2} = 1-\cos x = 1+\dfrac{4}{5} = \dfrac{9}{5}$, so $\sin^2\dfrac{x}{2} = \dfrac{9}{10}$, giving $\sin\dfrac{x}{2} = \dfrac{3}{\sqrt{10}}$ (positive, matching Step 1). Similarly $2\cos^2\dfrac{x}{2} = 1+\cos x = 1-\dfrac{4}{5} = \dfrac{1}{5}$, so $\cos^2\dfrac{x}{2} = \dfrac{1}{10}$, giving $\cos\dfrac{x}{2} = -\dfrac{1}{\sqrt{10}}$ (negative, again matching Step 1 — take the negative root, not the positive one, precisely because Step 1 already fixed the quadrant).

Step 4. Divide to get the tangent: $\tan\dfrac{x}{2} = \dfrac{\sin(x/2)}{\cos(x/2)} = \dfrac{3/\sqrt{10}}{-1/\sqrt{10}} = -3$, negative as predicted. Answer: $\sin\dfrac{x}{2} = \mathbf{\dfrac{3}{\sqrt{10}}}$, $\cos\dfrac{x}{2} = \mathbf{-\dfrac{1}{\sqrt{10}}}$, $\tan\dfrac{x}{2} = \mathbf{-3}$. Timing note: budget about 3 minutes. The entire problem hinges on one habit — write down the quadrant of the half-angle before you compute a single square root. Skipping Step 1 is the single most common way marks are lost on half-angle problems: the algebra of Steps 2–3 is unaffected by the quadrant, but the final $\pm$ sign is decided by nothing else, so a student who forgets to check it gets the magnitude right and the sign wrong on two of three answers.

The most frequent error is mixing degree and radian measure in one calculation — feeding a degree value into a formula that expects radians, especially the arc-length formula $l=r\theta$, valid only with $\theta$ in radians. A second, related trap is forgetting periodicity and trying to evaluate something like $\sin\dfrac{31\pi}{3}$ directly instead of first subtracting the largest possible multiple of $2\pi$, as NCERT's Example 8 shows. A third, very common trap is taking the "positive" root by default whenever a Pythagorean identity is inverted — $\sin x=\pm\sqrt{1-\cos^2x}$ always carries a genuine $\pm$, read off from the given quadrant, never assumed. A fourth, specific to half-angle problems, is finding the quadrant of $x$ but forgetting to work out the quadrant of $x/2$ separately, since they are often different (as in the competitive-level example above). A fifth is misremembering the sign pattern in the tangent compound-angle formulas — the denominator sign in $\tan(x\pm y)$ is always the opposite of the sign inside the bracket. A sixth is applying the sum-to-product formulas (13)–(14) with $x$ and $y$ swapped, which silently flips the sign of the antisymmetric results $\cos x-\cos y$ and $\sin x-\sin y$.

• Convert confidently between degree measure and radian measure in both directions, and apply $l=r\theta$ correctly with $\theta$ strictly in radians.
• State the unit-circle definition of $\cos x$ and $\sin x$ for any real $x$, and explain why it is more general than the right-triangle definition from earlier classes.
• Derive the four remaining functions ($\tan x,\cot x,\sec x,\text{cosec}\ x$) from $\sin x$ and $\cos x$, and state their domains precisely, including the excluded points.
• Determine the sign of any trigonometric function in any of the four quadrants using the ASTC rule, without hesitation.
• Find all six trigonometric functions given one ratio and the quadrant, correctly resolving every $\pm$ sign from the stated quadrant.
• Reproduce the geometric proof of $\cos(x+y)=\cos x\cos y-\sin x\sin y$ from the unit circle and the distance formula, and derive $\cos(x-y)$, $\sin(x\pm y)$ and $\tan(x\pm y)$ from it by substitution.
• Derive and correctly apply the double-angle formulas (all three forms of $\cos 2x$) and the triple-angle formulas for $\sin 3x$ and $\cos 3x$.
• Convert between sums/differences and products of sines and cosines using the sum-to-product and product-to-sum formulas, and recognise when a problem calls for one direction over the other.
• Handle half-angle problems correctly by determining the quadrant of $x/2$ separately from the quadrant of $x$, before resolving any $\pm$ sign.
• Recognise and avoid the chapter's classic traps: degree-radian mixing, ignoring periodicity, defaulting to the positive root, and misremembering the tangent formula's sign pattern.