Trigonometric Functions

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 3.

How far does the tip of a minute hand travel in 20 minutes?

Take a minute hand 10 cm long. In 20 minutes its tip moves along a piece of a circle, and you can find that distance with one multiplication, as long as you measure the turn in the right unit. In Class X, trigonometry was about right triangles and acute angles. Here it becomes something much bigger: sine and cosine are defined for every real number, negative, huge or fractional, and they describe anything that repeats, from a spinning wheel to an alternating current. The answer to the minute hand comes in the first worked example.

Measuring angles: degrees and radians

An angle is an amount of rotation. A ray turns about its starting point, called the vertex, from an initial side to a terminal side. Anticlockwise rotation is positive and clockwise rotation is negative. A turn can be larger than one full revolution, so angles like 420° or −30° make sense.

A degree is 1/360 of a full turn, with $1^\circ = 60'$ and $1' = 60''$. So $30^\circ 30' = 30.5^\circ$, not 30.30°.

A radian is a more natural unit. Draw a circle of radius r and cut off an arc of length l. The angle at the centre depends only on the ratio $l / r$, because a bigger circle gives a bigger arc but the same ratio. One radian is the angle when $l = r$, that is, when the arc is exactly as long as the radius. In general

$$\theta = \dfrac{l}{r} \quad \Longrightarrow \quad l = r\theta \qquad (\theta \text{ in radians})$$

A full turn has arc length $2\pi r$, so it equals $2\pi$ radians. That gives the conversion you must know cold:

$$\pi \text{ radian} = 180^\circ, \qquad \text{radians} = \text{degrees} \times \dfrac{\pi}{180}, \qquad \text{degrees} = \text{radians} \times \dfrac{180}{\pi}$$

One radian is about $57.3^\circ$ (it is $57^\circ 16'$ if you take $\pi = 22/7$), and one degree is about $0.0175$ radian.

Radians also connect angles to the whole real line. Imagine wrapping the number line around the unit circle, with 0 at the starting point. The number $x$ lands at the point where the rotation is $x$ radians, positive numbers going anticlockwise and negative numbers clockwise. This is why trigonometric functions accept any real number as input.

Worked example: the minute hand. A minute hand 10 cm long runs for 20 minutes. How far does its tip travel?

Common wrong step. Putting $\theta = 120$ (degrees) straight into $l = r\theta$, which gives 1200 cm. The formula works only when the angle is in radians.

Pause and predict: is $\sin 30$ the same as $\sin 30^\circ$?

Check your answer: no. With no degree sign the 30 means 30 radians. Since $30 - 8\pi \approx 4.87$, which lies in the fourth quadrant where sine is negative, $\sin 30 \approx -0.99$, while $\sin 30^\circ = 0.5$. This is also why calculators have a degree and a radian mode.

The unit circle: sine and cosine as coordinates

Draw the circle of radius 1 centred at the origin. Rotate from the positive x-axis through x radians and land at a point $P(a, b)$. Then

$$\cos x = a, \qquad \sin x = b$$

Because P is on the unit circle, $a^2 + b^2 = 1$, which gives the identity everything else rests on:

$$\sin^2 x + \cos^2 x = 1$$

The other four functions are built from these two. The identities with tan and cot follow by dividing the main identity by $\cos^2 x$ or $\sin^2 x$:

$$\tan x = \dfrac{\sin x}{\cos x}, \quad \cot x = \dfrac{\cos x}{\sin x}, \quad \sec x = \dfrac{1}{\cos x}, \quad \operatorname{cosec} x = \dfrac{1}{\sin x}$$

$$1 + \tan^2 x = \sec^2 x, \qquad 1 + \cot^2 x = \operatorname{cosec}^2 x$$

Reading the coordinates of the four points where the circle meets the axes gives the quadrantal values:

For the standard acute angles there is a pattern worth learning. The values of $\sin$ at $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ are $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$, and $\cos$ is the same list in reverse.

Signs by quadrant. The sign of each function is the sign of the coordinates of P. The point P has x-coordinate $\cos x$ and y-coordinate $\sin x$, so the signs follow from where P lies.

Remember it as All, Sin, Tan, Cos going anticlockwise from Quadrant I.

Domain, range and period. After a full turn of $2\pi$ you are back at the same point, so sine and cosine repeat every $2\pi$. For tangent, the points at x and $x + \pi$ are opposite points of the circle, so both coordinates change sign and the ratio does not change. That is why tan and cot repeat every $\pi$. The function sine is odd and cosine is even, because reflecting the point in the x-axis keeps $a$ and flips $b$:

$$\sin(-x) = -\sin x, \qquad \cos(-x) = \cos x$$

Here n is any integer. Also $\sin x = 0$ exactly when $x = n\pi$, and $\cos x = 0$ exactly when $x = (2n+1)\dfrac{\pi}{2}$.

Worked example: finding all six from one. If $\sin x = \dfrac{3}{5}$ and x lies in the second quadrant, find $\cos x$, $\tan x$ and $\sec x$.

Common wrong step. Writing $\cos x = +\dfrac{4}{5}$ after taking the square root. The root gives only the size. The quadrant decides the sign.

Worked example: reducing big angles. Find (a) $\sin\dfrac{31\pi}{3}$ and (b) $\cos(-1710^\circ)$.

Common wrong step. In (a), subtracting $9\pi$ instead of $10\pi$. That leaves $\sin\dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}$, with the wrong sign. For sine and cosine you may remove only multiples of $2\pi$. Odd multiples of $\pi$ change the sign.

Adding and subtracting angles

Everything below follows from one formula, which you can prove in a few lines. Put two points on the unit circle, $P(\cos x, \sin x)$ and $Q(\cos y, \sin y)$. The squared distance between them is

$$PQ^2 = (\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 2 - 2(\cos x \cos y + \sin x \sin y)$$

using $\sin^2 + \cos^2 = 1$ twice. Now turn the whole picture clockwise through y. The point Q lands on $(1, 0)$ and P lands at angle $x - y$, so $PQ^2 = 2 - 2\cos(x - y)$. A turn does not change distances, so the two expressions agree:

$$\cos(x - y) = \cos x \cos y + \sin x \sin y$$

Replace y by $-y$ to get the version with a plus. Putting $x = \dfrac{\pi}{2}$ shows $\cos\left(\dfrac{\pi}{2} - y\right) = \sin y$, and that gives the sine formulas. The whole family:

Notice that cosine flips the sign between left and right, while sine keeps it. Special cases of these formulas give the reduction rules, so you never need to memorise them separately:

Worked example: two uses of the formulas. (a) Find $\sin 75^\circ$. (b) If $\tan x = \dfrac{1}{2}$ and $\tan y = \dfrac{1}{3}$ with x and y acute, find $x + y$.

Common wrong step. Writing $\sin(A + B) = \sin A + \sin B$. A quick sanity check kills it: $\sin 45^\circ + \sin 30^\circ \approx 1.21$ is more than 1, which no sine can be. In (b), another slip is using $1 + \tan x \tan y$ in the denominator, which gives the formula for $\tan(x - y)$.

Pause and predict: is $\cos(x + y) = \cos x + \cos y$? Try $x = y = 60^\circ$.

Check your answer: no. The left side is $\cos 120^\circ = -\dfrac{1}{2}$, and the right side is $\dfrac{1}{2} + \dfrac{1}{2} = 1$.

Double, triple and half angles

Put $y = x$ in the sum formulas and the double-angle formulas appear at once:

$$\sin 2x = 2\sin x \cos x = \dfrac{2\tan x}{1 + \tan^2 x}, \qquad \tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}$$

$$\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x = \dfrac{1 - \tan^2 x}{1 + \tan^2 x}$$

The three forms of $\cos 2x$ are linked by $\sin^2 x + \cos^2 x = 1$. Choose the one that leaves you with the quantity you know. Writing $3x = 2x + x$ and using the double-angle results gives the triple angle formulas:

$$\sin 3x = 3\sin x - 4\sin^3 x, \qquad \cos 3x = 4\cos^3 x - 3\cos x, \qquad \tan 3x = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$$

For $\sin 3x$ the steps are $\sin 2x \cos x + \cos 2x \sin x = 2\sin x \cos^2 x + (1 - 2\sin^2 x)\sin x = 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x = 3\sin x - 4\sin^3 x$.

Solving the third form of $\cos 2x$ for $\sin^2 x$ and the second for $\cos^2 x$, and then writing $\dfrac{x}{2}$ in place of x, gives the half-angle formulas:

$$\sin^2\dfrac{x}{2} = \dfrac{1 - \cos x}{2}, \qquad \cos^2\dfrac{x}{2} = \dfrac{1 + \cos x}{2}$$

After taking a square root, the sign is that of $\sin\frac{x}{2}$ or $\cos\frac{x}{2}$ in the quadrant where $\frac{x}{2}$ lies, not where x lies.

Worked example: double angle from one value. If $\sin x = \dfrac{3}{5}$ and x is acute, find $\sin 2x$, $\cos 2x$ and $\tan 2x$.

Common wrong step. Writing $\sin 2x = 2\sin x = \dfrac{6}{5}$. Sine can never exceed 1, so the answer is impossible. The factor 2 belongs to the product $\sin x \cos x$, not to $\sin x$ alone.

Pause and predict: what does $3\sin x - 4\sin^3 x$ equal at $x = 30^\circ$?

Check your answer: $3 \cdot \dfrac{1}{2} - 4 \cdot \dfrac{1}{8} = \dfrac{3}{2} - \dfrac{1}{2} = 1$, which is $\sin 90^\circ = \sin 3x$. Putting in a known angle is the fastest way to catch a misremembered triple angle formula.

Sums into products, and back

Add the formulas for $\sin(x + y)$ and $\sin(x - y)$ and the cross terms cancel, leaving $2\sin x \cos y = \sin(x + y) + \sin(x - y)$. The other three product forms come from the same trick with the cosine formulas. Now write $C = x + y$ and $D = x - y$, so that $x = \dfrac{C + D}{2}$ and $y = \dfrac{C - D}{2}$, and read the identities backwards:

Only $\cos C - \cos D$ carries a minus sign out front.

Worked example: simplify a ratio. Simplify $\dfrac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x}$.

Common wrong step. Replacing $\sin 7x - \sin 5x$ by $\sin 2x$. The sine of a difference is not the difference of sines. A quick test with $x = 30^\circ$ makes this clear: $\sin 210^\circ - \sin 150^\circ = -1$, but $\sin 60^\circ \approx 0.87$.

The same method gives exact values, for example $\sin 75^\circ + \sin 15^\circ = 2\sin 45^\circ \cos 30^\circ = \dfrac{\sqrt{6}}{2}$.

Beyond the textbook. For the exam: convert degrees to radians whenever a formula such as $l = r\theta$ needs it. In reality: the radian is not just convenient, it is the unit in which calculus is clean. The slope of $\sin x$ is exactly $\cos x$, and $\dfrac{\sin x}{x}$ tends to 1 as x tends to 0, only when x is in radians. In degrees every such formula picks up an extra factor of $\dfrac{\pi}{180}$. Why it matters: you will meet this in Limits and Derivatives, and it is the reason this chapter makes you convert so carefully.

Beyond the textbook. For the exam: sine and cosine are coordinates on a circle. In reality: the same functions describe anything that repeats smoothly. Mains electricity is a voltage that follows $V_0 \sin(\omega t)$, and in India it repeats 50 times a second, so $\omega = 100\pi$ radians per second. Sound waves, tides and the swing of a pendulum for small angles follow the same shape. Why it matters: period $2\pi$ is not an abstract fact, it is the length of one full cycle of a real signal.

Beyond the textbook. For the exam: the six functions are defined from the unit circle. In reality: the word "sine" has a travelling history. Indian astronomers such as Aryabhata, around 500 CE, tabulated the half chord of an arc, which they called jya. Arabic writers borrowed it as jiba, it was misread as jayb (a fold or pocket), and the Latin translators rendered that as sinus. Why it matters: it explains why the "sine" is a half chord and why trigonometry grew out of astronomy.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Relations and Functions taught domain, range and graphs, and the six functions here are the next real functions to study with those tools.
• Next: Complex Numbers and Quadratic Equations puts numbers on a plane, where multiplying by $i$ is a quarter turn, the $\pi/2$ rotation you have just met.
• Straight lines: Straight Lines uses $\tan\theta$ as the slope of a line, together with $\tan(x \pm y)$ for the angle between two lines.
• Class XII link: Inverse Trigonometric Functions restricts the domains of these functions so that each can be reversed.