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Trigonometric Functions

Angles and Radian Measure

An angle is the rotation of a ray about its starting point (the vertex). The ray's starting position is the initial side and its final position the terminal side. Rotation anticlockwise gives a positive angle; clockwise gives a negative one. Trigonometry needs angles far beyond the $0^\circ$ to $90^\circ$ of a triangle, so we measure rotation itself, in two standard units.

In degree measure one full turn is $360^\circ$, and $1^\circ$ is split into $60$ minutes ($60'$) and each minute into $60$ seconds ($60''$). The unit is convenient but arbitrary — there is nothing special about the number $360$.

In radian measure the angle is the ratio of arc length to radius, which makes it a pure number. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius:

$$\theta = \dfrac{\text{arc length}}{\text{radius}} = \dfrac{l}{r} \quad\Rightarrow\quad l = r\theta \ \ (\theta \text{ in radians})$$

Since a full circle has circumference $2\pi r$, a complete turn is $\dfrac{2\pi r}{r} = 2\pi$ radians. Equating the two units for a half-turn gives the single conversion fact you must never forget:

$$\pi \text{ radians} = 180^\circ \quad\Rightarrow\quad 1^\circ = \dfrac{\pi}{180}\text{ rad}, \qquad 1 \text{ rad} = \dfrac{180^\circ}{\pi} \approx 57^\circ 16'$$

The common angles, in both units, are worth memorising as a block:

Degrees$30^\circ$$45^\circ$$60^\circ$$90^\circ$$180^\circ$$360^\circ$
Radians$\dfrac{\pi}{6}$$\dfrac{\pi}{4}$$\dfrac{\pi}{3}$$\dfrac{\pi}{2}$$\pi$$2\pi$

The relation $l = r\theta$ also gives the area of a sector of angle $\theta$ (in radians): $A = \dfrac{1}{2} r^2 \theta = \dfrac{1}{2} l r$. To convert, multiply degrees by $\dfrac{\pi}{180}$, or radians by $\dfrac{180}{\pi}$.

Deeper Insight — why radians, not degrees, are the natural unit: The degree is a human invention; the radian is forced on us by geometry. Because a radian is defined as a ratio of two lengths, it has no units at all, and that single fact is what makes the clean formulas $l = r\theta$ and $A = \tfrac{1}{2}r^2\theta$ possible — neither would hold if $\theta$ were in degrees, since you would have to bury an extra factor of $\tfrac{\pi}{180}$ inside. The deeper payoff comes in calculus: the result $\dfrac{d}{dx}\sin x = \cos x$ is true only when $x$ is in radians, because it secretly relies on $\lim_{\theta \to 0} \tfrac{\sin\theta}{\theta} = 1$, a limit that equals $1$ in radians and $\tfrac{\pi}{180}$ in degrees. So when a question gives no degree symbol, the number is in radians by default. Treat $\pi = 180^\circ$ as a unit conversion, exactly like metres to centimetres, and the whole topic becomes bookkeeping rather than memory.

One radian defined by an arc equal to the radius One Radian: arc length = radius θ=1 r arc = r A full turn= 2π rad= 360°π rad = 180° Arc length and sector area for angle theta Arc l = rθ and Sector area = ½ r²θ r l = rθ θ Degrees × π/180 → radRadians × 180/π → deg
Example 1: Convert $40^\circ$ into radian measure.
  1. Multiply the degree value by $\dfrac{\pi}{180}$.
  2. $40^\circ = 40 \times \dfrac{\pi}{180} = \dfrac{40\pi}{180}$ rad.
  3. Simplify the fraction: $\dfrac{40}{180} = \dfrac{2}{9}$.

Answer: $40^\circ = \dfrac{2\pi}{9}$ radians.

Example 2: Convert $\dfrac{5\pi}{6}$ radians into degree measure.
  1. Multiply the radian value by $\dfrac{180}{\pi}$.
  2. $\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{6}$.
  3. $= \dfrac{900}{6} = 150$.

Answer: $\dfrac{5\pi}{6}$ rad $= 150^\circ$.

Example 3: Find the length of an arc of a circle of radius $14$ cm that subtends an angle of $\dfrac{\pi}{4}$ at the centre.
  1. Use $l = r\theta$ with $\theta$ already in radians.
  2. $l = 14 \times \dfrac{\pi}{4} = \dfrac{14\pi}{4} = \dfrac{7\pi}{2}$ cm.
  3. Taking $\pi \approx \dfrac{22}{7}$: $l \approx \dfrac{7}{2} \times \dfrac{22}{7} = 11$ cm.

Answer: $l = \dfrac{7\pi}{2}$ cm $\approx 11$ cm.

Example 4: A wheel makes $360$ revolutions in one minute. Through how many radians does it turn in one second?
  1. $360$ revolutions per minute $= \dfrac{360}{60} = 6$ revolutions per second.
  2. Each revolution is $2\pi$ radians.
  3. So in one second it turns $6 \times 2\pi = 12\pi$ radians.

Answer: $12\pi$ radians per second.

Example 5: The minute hand of a clock is $1.5$ cm long. How far does its tip move in $40$ minutes?
  1. In $60$ minutes the minute hand sweeps a full turn, $2\pi$ rad, so in $40$ minutes it sweeps $\theta = \dfrac{40}{60}\times 2\pi = \dfrac{4\pi}{3}$ rad.
  2. The tip traces an arc of radius $r = 1.5$ cm, so $l = r\theta$.
  3. $l = 1.5 \times \dfrac{4\pi}{3} = 2\pi$ cm.
  4. Numerically $l \approx 2 \times 3.14 = 6.28$ cm.

Answer: The tip moves $2\pi \approx 6.28$ cm.

Example 6: Two arcs of the same length subtend angles of $60^\circ$ and $75^\circ$ at the centres of two circles. Find the ratio of their radii.
  1. Convert: $60^\circ = \dfrac{\pi}{3}$ rad and $75^\circ = \dfrac{5\pi}{12}$ rad.
  2. Equal arcs means $r_1\theta_1 = r_2\theta_2$, so $\dfrac{r_1}{r_2} = \dfrac{\theta_2}{\theta_1}$.
  3. $\dfrac{r_1}{r_2} = \dfrac{5\pi/12}{\pi/3} = \dfrac{5\pi}{12}\times\dfrac{3}{\pi} = \dfrac{15}{12} = \dfrac{5}{4}$.

Answer: $r_1 : r_2 = 5 : 4$.

Quick recap
  • Anticlockwise rotation is positive, clockwise is negative; trigonometry measures rotation, not just triangle angles.
  • The master conversion is $\pi$ rad $= 180^\circ$: multiply degrees by $\dfrac{\pi}{180}$, radians by $\dfrac{180}{\pi}$.
  • A radian is a pure ratio $\theta = \dfrac{l}{r}$, so arc length is simply $l = r\theta$ (with $\theta$ in radians).
  • Sector area $= \dfrac{1}{2}r^2\theta = \dfrac{1}{2}lr$; a full turn is $2\pi$ rad $= 360^\circ$.
  • When no degree symbol is written, the angle is assumed to be in radians.
✓ Quick check
Express cos 2x in terms of tan x.
cos 2x = cos²x − sin²x. Rewriting as (cos²x − sin²x) / (cos²x + sin²x). Dividing numerator and denominator by cos²x gives (1 − tan²x) / (1 + tan²x).
If θ + φ = 90°, then sin θ equals:
Cofunction identity.

Fundamental Identities

The six trigonometric functions are defined for any angle using the unit circle (radius $1$, centred at the origin). If the terminal side of angle $\theta$ meets the circle at the point $P(x, y)$, then:

$$\cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \dfrac{y}{x}\ (x \ne 0)$$

The remaining three are reciprocals: $\csc\theta = \dfrac{1}{\sin\theta}$, $\sec\theta = \dfrac{1}{\cos\theta}$, $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$. Because $x$ and $y$ never exceed $1$ in magnitude on the unit circle, $-1 \le \sin\theta \le 1$ and $-1 \le \cos\theta \le 1$ always.

The sign of each function depends only on the signs of $x$ and $y$ in that quadrant — remembered as "All Silver Tea Cups" (All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4):

QuadrantPositive functions$\sin$$\cos$$\tan$
I ($0$ to $90^\circ$)all$+$$+$$+$
II ($90$ to $180^\circ$)$\sin,\csc$$+$$-$$-$
III ($180$ to $270^\circ$)$\tan,\cot$$-$$-$$+$
IV ($270$ to $360^\circ$)$\cos,\sec$$-$$+$$-$

From $x^2 + y^2 = 1$ on the unit circle come the three Pythagorean identities:

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

The sum and difference formulae let you break a compound angle apart:

$$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$$
$$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$$
$$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$$

Setting $B = A$ gives the double-angle formulae, with three equivalent forms for $\cos 2A$:

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

The last two forms of $\cos 2A$ rearrange into the half-angle identities $\sin^2 A = \dfrac{1 - \cos 2A}{2}$ and $\cos^2 A = \dfrac{1 + \cos 2A}{2}$.

Deeper Insight — one definition generates the entire formula sheet: Students often try to memorise dozens of identities as separate facts, but they all descend from a single source — the unit-circle point $(\cos\theta, \sin\theta)$. The Pythagorean identities are literally just $x^2 + y^2 = 1$ rewritten, and dividing that one equation by $\cos^2\theta$ or $\sin^2\theta$ produces the other two for free. The double-angle formulae are not new either: they are the sum formulae with $B$ replaced by $A$, and the three faces of $\cos 2A$ come from substituting $\sin^2 = 1 - \cos^2$ into one another. Even the signs across quadrants are not arbitrary rules to cram — they simply read off whether $x$ and $y$ are positive or negative where the terminal side lands. If you internalise the unit circle and the two sum formulae, you can reconstruct the rest in seconds under exam pressure, which is far safer than recalling a memorised list and hoping you got a sign right.

Unit circle defining cosine and sine as coordinates Unit Circle: P = (cosθ, sinθ) θ P cosθ sinθ ASTC Q1: all +Q2: sin +Q3: tan +Q4: cos + Graphs of sine and cosine over one period y = sin x (orange) and y = cos x (blue) π2π1−1
Example 1: Given $\sin\theta = \dfrac{3}{5}$ and $\theta$ lies in the second quadrant, find $\cos\theta$ and $\tan\theta$.
  1. Use $\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}$, so $\cos\theta = \pm\dfrac{4}{5}$.
  2. In Quadrant II cosine is negative, so $\cos\theta = -\dfrac{4}{5}$.
  3. $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$.

Answer: $\cos\theta = -\dfrac{4}{5}$, $\tan\theta = -\dfrac{3}{4}$.

Example 2: Find the exact value of $\cos 15^\circ$.
  1. Write $15^\circ = 45^\circ - 30^\circ$ and use $\cos(A - B) = \cos A\cos B + \sin A\sin B$.
  2. $\cos 15^\circ = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ$.
  3. $= \dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}}\cdot\dfrac{1}{2} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}}$.
  4. Rationalise: $\dfrac{\sqrt{3} + 1}{2\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

Answer: $\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

Example 3: Prove that $\dfrac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta$.
  1. Use the Pythagorean identities: numerator $= \sec^2\theta$, denominator $= \csc^2\theta$.
  2. $\dfrac{\sec^2\theta}{\csc^2\theta} = \dfrac{1/\cos^2\theta}{1/\sin^2\theta} = \dfrac{\sin^2\theta}{\cos^2\theta}$.
  3. $= \tan^2\theta$, which equals the right-hand side.

Answer: Identity proved: both sides equal $\tan^2\theta$.

Example 4: If $\tan A = \dfrac{1}{2}$ and $\tan B = \dfrac{1}{3}$, find $\tan(A + B)$.
  1. Apply $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$.
  2. Numerator: $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$.
  3. Denominator: $1 - \dfrac{1}{2}\cdot\dfrac{1}{3} = 1 - \dfrac{1}{6} = \dfrac{5}{6}$.
  4. $\tan(A + B) = \dfrac{5/6}{5/6} = 1$.

Answer: $\tan(A + B) = 1$ (so $A + B = 45^\circ$).

Example 5: If $\cos\theta = \dfrac{3}{5}$ with $\theta$ in the first quadrant, find $\sin 2\theta$ and $\cos 2\theta$.
  1. In Q1, $\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}$.
  2. $\sin 2\theta = 2\sin\theta\cos\theta = 2\cdot\dfrac{4}{5}\cdot\dfrac{3}{5} = \dfrac{24}{25}$.
  3. $\cos 2\theta = 2\cos^2\theta - 1 = 2\cdot\dfrac{9}{25} - 1 = \dfrac{18}{25} - 1 = -\dfrac{7}{25}$.

Answer: $\sin 2\theta = \dfrac{24}{25}$, $\cos 2\theta = -\dfrac{7}{25}$.

Example 6: Prove that $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$.
  1. Rewrite the numerator with the double-angle formula: $\sin 2\theta = 2\sin\theta\cos\theta$.
  2. Rewrite the denominator using $\cos 2\theta = 2\cos^2\theta - 1$, so $1 + \cos 2\theta = 2\cos^2\theta$.
  3. $\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta} = \dfrac{\sin\theta}{\cos\theta}$.
  4. $= \tan\theta$, matching the right-hand side.

Answer: Identity proved: the expression simplifies to $\tan\theta$.

Quick recap
  • On the unit circle $\cos\theta = x$, $\sin\theta = y$; hence $-1 \le \sin\theta, \cos\theta \le 1$.
  • Signs by quadrant follow "All Silver Tea Cups": all $+$ in Q1, only $\sin$ in Q2, only $\tan$ in Q3, only $\cos$ in Q4.
  • Pythagorean identities: $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$.
  • Sum/difference: $\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$.
  • Double angle: $\sin 2A = 2\sin A\cos A$ and $\cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1$.
✓ Quick check
What is the domain of the trigonometric function f(x) = sec x?
sec x = 1 / cos x. It is undefined wherever cos x = 0. The cosine function is zero at odd multiples of π/2. Therefore, the domain is R − {(2n+1)π/2}.
Which graph has vertical asymptotes at x = π/2 + nπ?
tan x is undefined at π/2 + nπ.

Sum, Difference and Multiple Angles

The compound-angle formulae sin(A ± B) and cos(A ± B) generate the double-angle results sin 2A = 2 sinA cosA and cos 2A = cos²A − sin²A = 2cos²A − 1.

These convert sums to products and simplify equations and maxima of a sinx + b cosx, whose maximum is √(a² + b²).

Example 1: Find the maximum value of 3 sinx + 4 cosx.
√(3² + 4²) = 5.
Example 2: Evaluate cos 2θ if cosθ = 1/√2.
2(1/2) − 1 = 0.
Quick recap
  • Double angle: sin 2A = 2 sinA cosA, cos 2A = 2cos²A − 1.
  • Max of a sinx + b cosx is √(a² + b²).
✓ Quick check
The minute hand of a wall clock in Amit's house is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14)
In 60 minutes, the minute hand completes one revolution (2π radians). In 40 minutes, it turns θ = (40/60) × 2π = 4π/3 radians. Distance moved l = rθ = 1.5 × (4π/3) = 2π = 2 × 3.14 = 6.28 cm.
A carpenter in Punjab is building a triangular wooden frame. Two angles of the frame are 45° and 30°. What is the sine of the third angle?
The sum of angles in a triangle is 180°. Third angle = 180° − (45° + 30°) = 105°. sin 105° = sin(90° + 15°) = cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = (√3 + 1)/(2√2).
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