Trigonometry

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Intuition

Trigonometry is the mathematics of rotation and oscillation: The unit circle is the key, as you rotate around it, the $x$ -coordinate traces out $\cos\theta$ and the $y$ -coordinate traces out $\sin\theta$ . This connects circular motion to wave patterns, which is why trigonometry appears in physics (waves, SHM), engineering (signal processing), and music (sound waves).

Why it matters: Trigonometry is essential for any problem involving angles, periodic phenomena, or circular motion. It’s the bridge between geometry (shapes) and algebra (equations), and it’s the foundation for calculus (differentiation/integration of trig functions).

The key insight: The identity $\sin^2\theta + \cos^2\theta = 1$ is just Pythagoras’ theorem applied to the unit circle. Every point on the unit circle satisfies $x^2 + y^2 = 1$ , and since $x = \cos\theta$ and $y = \sin\theta$ , the identity follows immediately. This is why trig identities are really just geometric relationships in disguise.

Board Coverage

Board	Paper	Notes
AQA	Paper 1, 2	Basic trig in P1; compound/double angle, trig equations in P2
Edexcel	P1, P2	Similar split
OCR (A)	Paper 1, 2	Includes small angle approximations
CIE (9709)	P1, P2, P3	Trig functions in P1; identities and equations in P2; further trig in P3

1. The Unit Circle Definitions

Definition. Consider the unit circle (radius 1, centre at the origin). For an angle $\theta$ Measured anticlockwise from the positive $x$ -axis, the point on the circle is $(\cos\theta, \sin\theta)$ .

Definition.

\cos\theta = \mathrm{the } x\mathrm{-coordinate of the point on the unit circle at angle } \theta

\sin\theta = \mathrm{the } y\mathrm{-coordinate of the point on the unit circle at angle } \theta

\tan\theta = \frac{\sin\theta}{\cos\theta} \quad (\cos\theta \neq 0)

Intuition. These definitions extend the right-triangle definitions (SOH CAH TOA) to all angles, Not just those between 0° and 90°. The unit circle makes clear why $\sin$ and $\cos$ are periodic with period $2\pi$ .

1.1 Radian Measure

Definition. One radian is the angle subtended at the centre of a circle by an arc equal in Length to the radius.

\theta\mathrm{ (radians)} = \frac{\mathrm{arc length}}{r}

The full circle: $2\pi$ radians $= 360^\circ$ So $\pi$ radians $= 180^\circ$ .

Theorem (Arc length and sector area). For a sector of radius $r$ and angle $\theta$ (in Radians):

\mathrm{Arc length } s = r\theta

\mathrm{Sector area } A = \frac{1}{2}r^2\theta

Proof. By definition, $\theta = s/r$ So $s = r\theta$ . The sector is a fraction $\frac{\theta}{2\pi}$ of the full circle (area $\pi r^2$ ), so $A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{1}{2}r^2\theta$ . $\blacksquare$

2. Fundamental Identities

2.1 Pythagorean Identity

Theorem. For all $\theta \in \mathbb{R}$ :

\sin^2\theta + \cos^2\theta = 1

Proof. The point $(\cos\theta, \sin\theta)$ lies on the unit circle $x^2 + y^2 = 1$ .

\cos^2\theta + \sin^2\theta = 1 \quad \blacksquare

Corollary. Dividing by $\cos^2\theta$ (where $\cos\theta \neq 0$ ):

1 + \tan^2\theta = \sec^2\theta

Dividing by $\sin^2\theta$ (where $\sin\theta \neq 0$ ):

\cot^2\theta + 1 = \cosec^2\theta

3. Compound Angle Formulas

3.1 Sine of a Sum

Theorem. For all $A, B \in \mathbb{R}$ :

\sin(A + B) = \sin A \cos B + \cos A \sin B

Proof (using rotation matrices). Consider the rotation of the plane by angle $A$ followed by Rotation by angle $B$ . The combined rotation is by angle $A + B$ .

The rotation matrix by angle $\theta$ is:

R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

Matrix multiplication gives: $R(A)R(B) = R(A + B)$ .

\begin{pmatrix} \cos A & -\sin A \\ \sin A & \cos A \end{pmatrix}\begin{pmatrix} \cos B & -\sin B \\ \sin B & \cos B \end{pmatrix} = \begin{pmatrix} \cos(A+B) & -\sin(A+B) \\ \sin(A+B) & \cos(A+B) \end{pmatrix}

The $(2, 1)$ entry of the product is:

\sin A \cos B + \cos A \sin B = \sin(A + B) \quad \blacksquare

3.2 Cosine of a Sum

Theorem.

\cos(A + B) = \cos A \cos B - \sin A \sin B

Proof. From the $(1, 1)$ entry of the matrix product above:

\cos A \cos B - \sin A \sin B = \cos(A + B) \quad \blacksquare

3.3 Tangent of a Sum

Theorem.

\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

Proof.

\begin{aligned} \tan(A + B) &= \frac{\sin(A + B)}{\cos(A + B)} \\ &= \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B} \end{aligned}

Divide numerator and denominator by $\cos A \cos B$ :

= \frac{\tan A + \tan B}{1 - \tan A \tan B} \quad \blacksquare

3.4 Difference Formulas

Theorem.

\begin{aligned} \sin(A - B) &= \sin A \cos B - \cos A \sin B \\ \cos(A - B) &= \cos A \cos B + \sin A \sin B \\ \tan(A - B) &= \frac{\tan A - \tan B}{1 + \tan A \tan B} \end{aligned}

Proof. Replace $B$ with $-B$ in the sum formulas, using $\sin(-B) = -\sin B$ and $\cos(-B) = \cos B$ . $\blacksquare$

4. Double Angle Formulas

Setting $A = B$ in the compound angle formulas:

\begin{aligned} \sin 2A &= 2\sin A \cos A \\ \cos 2A &= \cos^2 A - \sin^2 A \\ \tan 2A &= \frac{2\tan A}{1 - \tan^2 A} \end{aligned}

Theorem. Three equivalent forms of $\cos 2A$ :

\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A

Proof. Using $\sin^2 A = 1 - \cos^2 A$ :

$\cos^2 A - \sin^2 A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1$

Using $\cos^2 A = 1 - \sin^2 A$ :

$\cos^2 A - \sin^2 A = (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A$ $\blacksquare$

Intuition. The double angle formulas express functions of $2A$ purely in terms of functions of $A$ . They are the algebraic backbone of many trigonometric manipulations and are essential for Integration.

5. Solving Trigonometric Equations

5.1 Basic Strategy

Use identities to reduce to a single trigonometric function.
Solve the resulting equation.
Find all solutions in the required interval.

5.2 Key Solutions

\begin{aligned} \sin\theta = a &\implies \theta = \arcsin(a) + 2n\pi \mathrm{ or } \pi - \arcsin(a) + 2n\pi \\ \cos\theta = a &\implies \theta = \pm\arccos(a) + 2n\pi \\ \tan\theta = a &\implies \theta = \arctan(a) + n\pi \end{aligned}

Example

Solve

2\sin^2\theta + 3\cos\theta - 3 = 0

for

0 \leq \theta < 2\pi

Using $\sin^2\theta = 1 - \cos^2\theta$ :

2(1 - \cos^2\theta) + 3\cos\theta - 3 = 0

-2\cos^2\theta + 3\cos\theta - 1 = 0

2\cos^2\theta - 3\cos\theta + 1 = 0

(2\cos\theta - 1)(\cos\theta - 1) = 0

$\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ .

$\cos\theta = 1 \implies \theta = 0$ .

Solutions: $\theta = 0, \frac{\pi}{3}, \frac{5\pi}{3}$ .

6. Small Angle Approximations

Theorem. For small $\theta$ (in radians):

\begin{aligned} \sin\theta &\approx \theta - \frac{\theta^3}{6} \\ \cos\theta &\approx 1 - \frac{\theta^2}{2} \\ \tan\theta &\approx \theta + \frac{\theta^3}{3} \end{aligned}

These follow from the Maclaurin series expansions (see Differentiation).

For the linear approximations (when $\theta$ is very small):

\sin\theta \approx \theta, \quad \cos\theta \approx 1, \quad \tan\theta \approx \theta

Intuition. Near the origin, the curves $y = \sin\theta$ and $y = \tan\theta$ are almost Indistinguishable from the line $y = \theta$ . The curve $y = \cos\theta$ is nearly flat at $y = 1$ .

7. Half-Angle Formulas

Theorem. For all $\theta$ :

\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}

\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}

\tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}

7.1 Derivation

Starting from $\cos 2A = 1 - 2\sin^2 A$ and substituting $A = \frac{\theta}{2}$ :

\cos\theta = 1 - 2\sin^2\frac{\theta}{2}

2\sin^2\frac{\theta}{2} = 1 - \cos\theta

\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}} \quad \blacksquare

Similarly, from $\cos 2A = 2\cos^2 A - 1$ :

\cos\theta = 2\cos^2\frac{\theta}{2} - 1

2\cos^2\frac{\theta}{2} = 1 + \cos\theta

\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}} \quad \blacksquare

7.2 Tangent Half-Angle Formulas

The tangent half-angle formulas avoid the ambiguity of the $\pm$ sign:

\tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta}

Proof. Using the double angle formulas:

\frac{\sin\theta}{1 + \cos\theta} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{1 + (2\cos^2\frac{\theta}{2} - 1)} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2} \quad \blacksquare

Similarly:

\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}

Proof. $\frac{1 - \cos\theta}{\sin\theta} = \frac{1 - (1 - 2\sin^2\frac{\theta}{2})}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2} \quad \blacksquare$

7.3 Sign Determination

The $\pm$ in the sine and cosine half-angle formulas depends on the quadrant of $\frac{\theta}{2}$ Not the quadrant of $\theta$ itself. Always determine which quadrant $\frac{\theta}{2}$ lies in Before choosing the sign.

Quadrant of $\frac{\theta}{2}$	$\sin\frac{\theta}{2}$	$\cos\frac{\theta}{2}$
I: $0 \lt \frac{\theta}{2} \lt \frac{\pi}{2}$	$+$	$+$
II: $\frac{\pi}{2} \lt \frac{\theta}{2} \lt \pi$	$+$	$-$
III: $\pi \lt \frac{\theta}{2} \lt \frac{3\pi}{2}$	$-$	$-$
IV: $\frac{3\pi}{2} \lt \frac{\theta}{2} \lt 2\pi$	$-$	$+$

Worked example

Find the exact value of

\sin\frac{\pi}{8}

Since $0 \lt \frac{\pi}{8} \lt \frac{\pi}{2}$ (first quadrant), $\sin\frac{\pi}{8} > 0$ So we Take the positive root.

\sin\frac{\pi}{8} = \sqrt{\frac{1 - \cos\frac{\pi}{4}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}

8. R-Addition Formula (Harmonic Form)

Theorem. For real numbers $a$ and $b$ :

a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)

Where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \dfrac{b}{a}$ .

8.1 Cosine Form

Alternatively, the same expression can be written as:

a\sin\theta + b\cos\theta = R\cos(\theta - \beta)

Where $R = \sqrt{a^2 + b^2}$ and $\tan\beta = \dfrac{a}{b}$ .

Both forms are equivalent; the choice between them is a matter of convenience depending on whether a Sine or cosine expansion is more natural for the problem at hand.

8.2 Derivation

Expand the right-hand side of the sine form:

R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha

Equating coefficients with $a\sin\theta + b\cos\theta$ :

R\cos\alpha = a, \qquad R\sin\alpha = b

Squaring and adding: $R^2\cos^2\alpha + R^2\sin^2\alpha = a^2 + b^2$ So $R^2 = a^2 + b^2$ and $R = \sqrt{a^2 + b^2}$ .

Dividing the second equation by the first: $\tan\alpha = \dfrac{b}{a}$ . $\blacksquare$

The cosine form is derived similarly by expanding $R\cos(\theta - \beta) = R\cos\theta\cos\beta + R\sin\theta\sin\beta$ and equating coefficients.

8.3 Applications: Maximum and Minimum

Since $-1 \leq \sin(\theta + \alpha) \leq 1$ :

-R \leq a\sin\theta + b\cos\theta \leq R

Maximum $= R = \sqrt{a^2 + b^2}$ Occurring when $\theta + \alpha = \frac{\pi}{2} + 2n\pi$ .
Minimum $= -R$ Occurring when $\theta + \alpha = \frac{3\pi}{2} + 2n\pi$ .

Example: Finding maximum and minimum

Find the maximum and minimum values of

3\sin\theta - 4\cos\theta

and the values of

\theta

at which they occur, for

0 \leq \theta \lt 2\pi

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

Here $a = 3$ and $b = -4$ . Writing $3\sin\theta - 4\cos\theta = 5\sin(\theta + \alpha)$ where $\tan\alpha = \dfrac{-4}{3}$ So $\alpha = -\arctan\dfrac{4}{3}$ .

Maximum $= 5$ when $\theta + \alpha = \frac{\pi}{2}$ :

\theta = \frac{\pi}{2} - \alpha = \frac{\pi}{2} + \arctan\frac{4}{3} \approx 2.214 \mathrm{ rad}

Minimum $= -5$ when $\theta + \alpha = \frac{3\pi}{2}$ :

\theta = \frac{3\pi}{2} - \alpha = \frac{3\pi}{2} + \arctan\frac{4}{3} \approx 5.356 \mathrm{ rad}

8.4 Solving Equations

The R-addition formula converts $a\sin\theta + b\cos\theta = k$ into $R\sin(\theta + \alpha) = k$ A Standard trigonometric equation.

Example: Solving an equation

Solve

\sin\theta + \cos\theta = 1

for

0 \leq \theta \lt 2\pi

$R = \sqrt{1 + 1} = \sqrt{2}$ , $\alpha = \arctan 1 = \frac{\pi}{4}$ .

\sqrt{2}\sin\!\left(\theta + \frac{\pi}{4}\right) = 1

\sin\!\left(\theta + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \sin\frac{\pi}{4}

\theta + \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi \quad \mathrm{or} \quad \theta + \frac{\pi}{4} = \frac{3\pi}{4} + 2n\pi

\theta = 2n\pi \quad \mathrm{or} \quad \theta = \frac{\pi}{2} + 2n\pi

For $0 \leq \theta \lt 2\pi$ : $\theta = 0$ or $\theta = \frac{\pi}{2}$ .

9. Trigonometric Identities: Proof Strategies

Proving trigonometric identities is a core exam skill. The following strategies cover the most Common approaches.

9.1 Strategy 1: Work with One Side

Start from the more complicated side and simplify it until it matches the simpler side. This avoids the logical error of assuming what you are trying to prove.

9.2 Strategy 2: Express Everything in Sine and Cosine

Replace $\tan$ , $\sec$ , $\csc$ , $\cot$ with their definitions in terms of $\sin$ and $\cos$ : $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , $\sec\theta = \frac{1}{\cos\theta}$ Etc.

9.3 Strategy 3: Use Known Identities

Apply the Pythagorean identity, compound angle, or double angle formulas to create simplifications.

9.4 Strategy 4: Multiply by the Conjugate

When you see expressions of the form $a \pm b$ in a denominator or numerator, multiply top and Bottom by the conjugate $a \mp b$ to produce a difference of squares.

Example 1: Strategy 2 (express in sin and cos)

Prove that

\cot A + \tan A = \dfrac{2}{\sin 2A}

\begin{aligned} \cot A + \tan A &= \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A} \\ &= \frac{\cos^2 A + \sin^2 A}{\sin A \cos A} \\ &= \frac{1}{\sin A \cos A} \\ &= \frac{2}{2\sin A \cos A} = \frac{2}{\sin 2A} \quad \blacksquare \end{aligned}

Example 2: Strategy 3 (use known identities)

Prove that

\sin 3A = 3\sin A - 4\sin^3 A

\begin{aligned} \sin 3A &= \sin(2A + A) \\ &= \sin 2A \cos A + \cos 2A \sin A \\ &= 2\sin A \cos^2 A + (1 - 2\sin^2 A)\sin A \\ &= 2\sin A(1 - \sin^2 A) + \sin A - 2\sin^3 A \\ &= 2\sin A - 2\sin^3 A + \sin A - 2\sin^3 A \\ &= 3\sin A - 4\sin^3 A \quad \blacksquare \end{aligned}

Example 3: Strategy 4 (multiply by conjugate)

Prove that

\dfrac{1}{\sec A + \tan A} = \sec A - \tan A

Multiply numerator and denominator by $\sec A - \tan A$ :

\begin{aligned} \frac{1}{\sec A + \tan A} \cdot \frac{\sec A - \tan A}{\sec A - \tan A} &= \frac{\sec A - \tan A}{\sec^2 A - \tan^2 A} \\ &= \frac{\sec A - \tan A}{1 + \tan^2 A - \tan^2 A} \\ &= \frac{\sec A - \tan A}{1} \\ &= \sec A - \tan A \quad \blacksquare \end{aligned}

Where we used $\sec^2 A = 1 + \tan^2 A$ so that $\sec^2 A - \tan^2 A = 1$ .

Example 4: Strategy 1 (work with one side)

Prove that

\dfrac{\cos 2A}{1 + \sin 2A} = \dfrac{\cos A - \sin A}{\cos A + \sin A}

Working from the LHS:

\begin{aligned} \frac{\cos 2A}{1 + \sin 2A} &= \frac{\cos^2 A - \sin^2 A}{1 + 2\sin A \cos A} \\ &= \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos^2 A + 2\sin A\cos A + \sin^2 A} \\ &= \frac{(\cos A - \sin A)(\cos A + \sin A)}{(\cos A + \sin A)^2} \\ &= \frac{\cos A - \sin A}{\cos A + \sin A} \quad \blacksquare \end{aligned}

10. Trigonometric Graphs

10.1 Key Features

Function	Period	Amplitude	Domain	Range
$\sin x$	$2\pi$	$1$	$\mathbb{R}$	$[-1, 1]$
$\cos x$	$2\pi$	$1$	$\mathbb{R}$	$[-1, 1]$
$\tan x$	$\pi$	undefined	$x \neq \frac{\pi}{2} + n\pi$	$\mathbb{R}$

Key values of $\sin$ and $\cos$ :

Angle	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$\sin$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$	$0$	$-1$	$0$
$\cos$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$	$-1$	$0$	$1$

10.2 Transformations

For the general form $y = A\sin(Bx + C) + D$ (and similarly for $\cos$ ):

$|A|$ is the amplitude (vertical stretch from the midline)
The period is $\dfrac{2\pi}{|B|}$
The phase shift is $-\dfrac{C}{B}$ (horizontal shift)
$D$ is the vertical shift (midline is $y = D$ )

For $\tan$ The period is $\dfrac{\pi}{|B|}$ and amplitude is not defined.

Use the sliders to adjust the amplitude, period, phase shift, and Vertical translation of the trigonometric functions. Observe how each parameter affects the graph of $y = A\sin(Bx + C) + D$ .

Worked example

Describe the key features of

y = 2\sin\!\left(2x - \frac{\pi}{3}\right)

for

0 \leq x \leq 2\pi

. Amplitude:

|A| = 2

So the range is

[-2, 2]

. Period:

\dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi

. Phase shift:

-\dfrac{C}{B} = -\dfrac{-\pi/3}{2} = \dfrac{\pi}{6}

(shift right by

\frac{\pi}{6}

). Key points. The first cycle begins at

x = \frac{\pi}{6}

(where the curve crosses the midline Upward). Subsequent key points within

[0, 2\pi]

: - Maximum at

x = \frac{5\pi}{12}

(value

2

) - Midline crossing (down) at

x = \frac{2\pi}{3}

- Minimum at

x = \frac{11\pi}{12}

(value

-2

) - Midline crossing (up) at

x = \frac{7\pi}{6}

Since the period is

\pi

The second cycle repeats with all

x

-values shifted by

\pi

: - Maximum at

x = \frac{17\pi}{12}

(value

2

) - Minimum at

x = \frac{23\pi}{12}

(value

-2

) $y$ -intercept:

y = 2\sin\!\left(-\frac{\pi}{3}\right) = -\sqrt{3} \approx -1.73

11. Problem Set

Problem 1. Prove that $\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$ .

Solution

\begin{aligned} \frac{\sin 2\theta}{1 + \cos 2\theta} &= \frac{2\sin\theta\cos\theta}{1 + (2\cos^2\theta - 1)} \\ &= \frac{2\sin\theta\cos\theta}{2\cos^2\theta} \\ &= \frac{\sin\theta}{\cos\theta} = \tan\theta \quad \blacksquare \end{aligned}

If you get this wrong, revise: Double angle formulas

Problem 2. Solve $\sin 2x = \sin x$ for $0 \leq x < 2\pi$ .

Solution

2\sin x\cos x = \sin x

\sin x(2\cos x - 1) = 0

$\sin x = 0 \implies x = 0, \pi$ .

$2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}$ .

Solutions: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$ .

If you get this wrong, revise: Solving trig equations

Problem 3. Express $4\sin\theta - 3\cos\theta$ in the form $R\sin(\theta - \alpha)$ Where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$ .

Solution

R\sin(\theta - \alpha) = R(\sin\theta\cos\alpha - \cos\theta\sin\alpha) = R\cos\alpha\sin\theta - R\sin\alpha\cos\theta

Comparing coefficients: $R\cos\alpha = 4$ and $R\sin\alpha = 3$ .

R^2 = 4^2 + 3^2 = 25 \implies R = 5

\tan\alpha = \frac{3}{4} \implies \alpha = \arctan\left(\frac{3}{4}\right) \approx 0.6435 \mathrm{ rad}

4\sin\theta - 3\cos\theta = 5\sin(\theta - 0.6435)

If you get this wrong, revise: Compound angle formulas

Problem 4. Find the exact value of $\sin 75^\circ$ .

Solution

\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30^\circ

= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}

If you get this wrong, revise: Compound angle formulas

Problem 5. Prove that $\frac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta$ .

Solution

\begin{aligned} \frac{1 - \cos 2\theta}{\sin 2\theta} &= \frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} \\ &= \frac{2\sin^2\theta}{2\sin\theta\cos\theta} \\ &= \frac{\sin\theta}{\cos\theta} = \tan\theta \quad \blacksquare \end{aligned}

If you get this wrong, revise: Double angle formulas

Problem 6. A sector of a circle of radius 8 cm has an angle of 1.2 radians. Find the arc length And the area of the sector.

Solution

Arc length:

s = r\theta = 8 \times 1.2 = 9.6

cm.

Area: $A = \frac{1}{2}r^2\theta = \frac{1}{2}(64)(1.2) = 38.4$ cm².

If you get this wrong, revise: Radian measure

Problem 7. Solve $3\cos^2 x + 2\sin x - 2 = 0$ for $-\pi \leq x \leq \pi$ .

Solution

3(1 - \sin^2 x) + 2\sin x - 2 = 0

$3 - 3\sin^2 x + 2\sin x - 2 = 0$

$-3\sin^2 x + 2\sin x + 1 = 0$

$3\sin^2 x - 2\sin x - 1 = 0$

$(3\sin x + 1)(\sin x - 1) = 0$

$\sin x = -\frac{1}{3} \implies x = \arcsin(-\frac{1}{3}) \approx -0.3398$ or $x = -\pi - \arcsin(-\frac{1}{3}) = -\pi + 0.3398 \approx -2.802$ (since $3.481 > \pi$ Subtract $2\pi$ to stay in $[-\pi, \pi]$ ).

$\sin x = 1 \implies x = \frac{\pi}{2}$ .

Solutions: $x \approx -2.802, -0.340, \frac{\pi}{2}$ .

If you get this wrong, revise: Solving trig equations

Problem 8. Use small angle approximations to estimate $\frac{\sin 0.05}{\cos 0.05}$ .

Solution

\sin 0.05 \approx 0.05

$\cos 0.05 \approx 1 - \frac{0.05^2}{2} = 1 - 0.00125 = 0.99875$

\frac{\sin 0.05}{\cos 0.05} \approx \frac{0.05}{0.99875} \approx 0.05006

(Alternatively, $\tan 0.05 \approx 0.05$ directly.)

If you get this wrong, revise: Small angle approximations

Problem 9. Prove that $\dfrac{1 + \sin 2A}{\cos 2A} = \tan\!\left(A + \dfrac{\pi}{4}\right)$ .

Solution

Working from the LHS:

\begin{aligned} \frac{1 + \sin 2A}{\cos 2A} &= \frac{1 + 2\sin A \cos A}{\cos^2 A - \sin^2 A} \\ &= \frac{\sin^2 A + 2\sin A \cos A + \cos^2 A}{(\cos A - \sin A)(\cos A + \sin A)} \\ &= \frac{(\sin A + \cos A)^2}{(\cos A - \sin A)(\cos A + \sin A)} \\ &= \frac{\sin A + \cos A}{\cos A - \sin A} \end{aligned}

Dividing numerator and denominator by $\cos A$ :

= \frac{\tan A + 1}{1 - \tan A}

From the tangent compound angle formula:

\tan\!\left(A + \frac{\pi}{4}\right) = \frac{\tan A + \tan\frac{\pi}{4}}{1 - \tan A \tan\frac{\pi}{4}} = \frac{\tan A + 1}{1 - \tan A} \quad \blacksquare

If you get this wrong, revise: Compound angle formulas

Problem 10. The area of a sector is $20\mathrm{ cm}^2$ and the arc length is $10\mathrm{ cm}$ . Find the radius and the angle.

Solution

s = r\theta = 10 \implies \theta = \frac{10}{r}

A = \frac{1}{2}r^2\theta = \frac{1}{2}r^2 \cdot \frac{10}{r} = 5r = 20 \implies r = 4

\theta = \frac{10}{4} = 2.5 \mathrm{ radians}

If you get this wrong, revise: Radian measure

Problem 11. Find the exact value of $\cos\dfrac{\pi}{12}$ using a half-angle formula.

Solution

\cos\frac{\pi}{12} = \cos\frac{\pi/6}{2}

So we apply the half-angle formula with

\theta = \frac{\pi}{6}

Since $0 \lt \frac{\pi}{12} \lt \frac{\pi}{2}$ (first quadrant), $\cos\frac{\pi}{12} > 0$ .

\begin\{aligned\} \cos\frac\{\pi\}\{12\} &= \sqrt\{\frac\{1 + \cos\frac\{\pi\}\{6\}\}\{2\}\} \\ &= \sqrt\{\frac\{1 + \frac\{\sqrt\{3\}\}\{2\}\}\{2\}\} \\ &= \sqrt\{\frac\{2 + \sqrt\{3\}\}\{4\}\} \\ &= \frac\{\sqrt\{2 + \sqrt\{3\}\}\}\{2\} \end\{aligned\}

This can also be verified using the compound angle formula: $\cos\frac{\pi}{12} = \cos\!\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\frac{\pi}{3}\cos\frac{\pi}{4} + \sin\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$ and one can check that $\dfrac{\sqrt{2+\sqrt{3}}}{2} = \dfrac{\sqrt{6}+\sqrt{2}}{4}$ .

If you get this wrong, revise: Half-angle formulas

Problem 12. Find the maximum value of $2\sin\theta + 5\cos\theta$ and the smallest positive Value of $\theta$ at which it occurs.

Solution

R = \sqrt{4 + 25} = \sqrt{29}

The maximum value is $R = \sqrt{29}$ .

Writing $2\sin\theta + 5\cos\theta = \sqrt{29}\sin(\theta + \alpha)$ where $\tan\alpha = \dfrac{5}{2}$ .

The maximum occurs when $\sin(\theta + \alpha) = 1$ I.e., $\theta + \alpha = \frac{\pi}{2}$ .

\theta = \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \arctan\frac{5}{2} \approx 0.3805 \mathrm{ rad}

Since $\arctan\frac{5}{2} \approx 1.1903 < \frac{\pi}{2}$ This $\theta$ is positive and is the Smallest positive value.

If you get this wrong, revise: R-addition formula

Problem 13. Prove that $\dfrac{\sin 3A}{\sin A} - \dfrac{\cos 3A}{\cos A} = 2$ .

Solution

\begin{aligned} \frac{\sin 3A}{\sin A} - \frac{\cos 3A}{\cos A} &= \frac{\sin 3A \cos A - \cos 3A \sin A}{\sin A \cos A} \\ &= \frac{\sin(3A - A)}{\sin A \cos A} \\ &= \frac{\sin 2A}{\sin A \cos A} \\ &= \frac{2\sin A \cos A}{\sin A \cos A} \\ &= 2 \quad \blacksquare \end{aligned}

If you get this wrong, revise: Proof strategies

Problem 14. Solve $\sin\theta + \sqrt{3}\cos\theta = 1$ for $0 \leq \theta \lt 2\pi$ .

Solution

R = \sqrt{1 + 3} = 2

\alpha = \arctan\sqrt{3} = \frac{\pi}{3}

2\sin\!\left(\theta + \frac{\pi}{3}\right) = 1

\sin\!\left(\theta + \frac{\pi}{3}\right) = \frac{1}{2}

\theta + \frac{\pi}{3} = \frac{\pi}{6} + 2n\pi \quad \mathrm{or} \quad \theta + \frac{\pi}{3} = \frac{5\pi}{6} + 2n\pi

Case 1: $\theta = \frac{\pi}{6} - \frac{\pi}{3} + 2n\pi = -\frac{\pi}{6} + 2n\pi$ .

For $n = 1$ : $\theta = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$ .

Case 2: $\theta = \frac{5\pi}{6} - \frac{\pi}{3} + 2n\pi = \frac{\pi}{2} + 2n\pi$ .

For $n = 0$ : $\theta = \frac{\pi}{2}$ .

Solutions: $\theta = \frac{\pi}{2}$ or $\theta = \frac{11\pi}{6}$ .

If you get this wrong, revise: R-addition formula

Problem 15. Prove that $\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos 2A$ .

Solution

Using

\sec^2 A = 1 + \tan^2 A

\begin{aligned} \frac{1 - \tan^2 A}{1 + \tan^2 A} &= \frac{1 - \tan^2 A}{\sec^2 A} \\ &= (1 - \tan^2 A)\cos^2 A \\ &= \cos^2 A - \tan^2 A \cos^2 A \\ &= \cos^2 A - \frac{\sin^2 A}{\cos^2 A} \cdot \cos^2 A \\ &= \cos^2 A - \sin^2 A \\ &= \cos 2A \quad \blacksquare \end{aligned}

If you get this wrong, revise: Proof strategies

Tip Ready to test your understanding of Trigonometry? The contains the hardest questions

within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Trigonometry with other pure mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Common Pitfalls

Rounding too early in multi-step calculations. Carry full precision through and round only the final answer.
Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).
Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.
Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.

flowchart TD
    A[08 Trigonometryx] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

Differentiation — Trigonometric derivatives and small angle approximations connect directly to differentiation techniques.
Integration — Trigonometric integrals and substitutions require fluency with the identities and substitutions covered here.
Further Pure Mathematics 1 — Complex numbers use modulus-argument form, which is built on the trigonometric functions defined in this topic.
Functions — Trigonometric functions are key examples of periodic, composite, and inverse functions.

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