Trigonometry

Trigonometry describes how angles, coordinates and lengths are connected. At A level, it grows from right angled triangle ratios into a study of periodic functions. The central definitions are most usefully understood on the unit circle:

$$\boxed{\cos\theta=x}, \qquad \boxed{\sin\theta=y}, \qquad \boxed{\tan\theta=\frac{y}{x}=\frac{\sin\theta}{\cos\theta}}$$

for the point $(x,y)$ reached by turning through an angle $\theta$ from the positive $x$ axis on the circle $x^2+y^2=1$.

This viewpoint explains negative values, angles larger than one full turn, exact values, identities and the repeating shapes of trigonometric graphs. It is more powerful than remembering sine, cosine and tangent only as SOHCAHTOA.

Prerequisites

You should be able to:

• use sine, cosine and tangent in right angled triangles;
• apply Pythagoras’ theorem and rearrange formulae;
• manipulate fractions, surds and simple algebraic expressions;
• solve linear and quadratic equations;
• interpret coordinates, functions and graph transformations;
• set a calculator to degrees or radians deliberately.

Review Pythagoras and trigonometry, surds and graphs of functions if these skills are insecure.

The unit circle unifies the subject

On the unit circle, the point corresponding to $\theta$ has coordinates

$$(\cos\theta,\sin\theta).$$

Since every point on the circle satisfies $x^2+y^2=1$,

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

Therefore the most important identity in elementary trigonometry is not an isolated fact. It is simply Pythagoras’ theorem on a circle:

$$\boxed{\sin^2\theta+\cos^2\theta\equiv1}.$$

The symbol $\equiv$ means that the statement is true for every value of $\theta$ for which both sides are defined. This differs from an equation such as $\sin\theta=\frac12$, which is true only for particular angles.

Worked example 1: use a quadrant and an identity

An angle $\theta$ lies in the second quadrant and

$$\sin\theta=\frac35.$$

Find $\cos\theta$ and $\tan\theta$ exactly.

From $\sin^2\theta+\cos^2\theta=1$,

$$\begin{aligned} \cos^2\theta &=1-\left(\frac35\right)^2\\ &=1-\frac9{25}\\ &=\frac{16}{25}. \end{aligned}$$

Thus $\cos\theta=\pm\frac45$. The second quadrant has negative $x$ coordinates, so cosine is negative:

$$\cos\theta=-\frac45.$$

Hence

$$\tan\theta =\frac{\sin\theta}{\cos\theta} =\frac{3/5}{-4/5} =-\frac34.$$

Therefore

$$\boxed{\cos\theta=-\frac45, \qquad \tan\theta=-\frac34}.$$

The square root produces two possible signs. The stated quadrant selects the correct one. Losing the $\pm$ before considering the quadrant is a common source of errors.

Recommended learning route

The lessons are ordered by mathematical dependency. If the early ideas are secure, move directly to the point where your knowledge becomes uncertain.

1. Exact values, angle measure and graphs

Begin with the language and geometry used everywhere else.

1. Exact trigonometric values derives the values for $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$, then extends them using symmetry and quadrants.
2. Radians, arcs and sectors replaces degree measure with the natural angle measure for circles and calculus.
3. Trigonometric graphs develops the shapes, periods, symmetries and transformations of sine, cosine and tangent.

Degrees measure a turn by dividing it into $360$ parts. Radians measure it by comparing arc length with radius:

$$\theta=\frac{s}{r}, \qquad s=r\theta,$$

where $\theta$ must be in radians in the second formula. Since a complete circumference is $2\pi r$,

$$360^\circ=2\pi\text{ radians}, \qquad 180^\circ=\pi\text{ radians}.$$

Worked example 2: read a transformation structurally

State the amplitude, period and range of

$$y=3\sin(2x)-1,$$

where $x$ is in radians.

For $y=a\sin(bx)+c$:

• $|a|$ is the amplitude;
• the period is $\frac{2\pi}{|b|}$;
• $c$ is the midline.

Here

$$|a|=3, \qquad \text{period}=\frac{2\pi}{2}=\pi, \qquad c=-1.$$

Since $-1\leq\sin(2x)\leq1$,

$$-3\leq3\sin(2x)\leq3,$$

and therefore

$$\boxed{-4\leq y\leq2}.$$

The vertical translation changes the midline and range, but not the amplitude. The factor $2$ inside the function halves the period.

2. Functions, identities and equations

Next, learn how trigonometric expressions behave algebraically.

4. Reciprocal and inverse trigonometric functions distinguishes $\sec x$, $\csc x$ and $\cot x$ from inverse functions such as $\sin^{-1}x$.
5. Trigonometric identities develops Pythagorean, quotient and reciprocal identities, then uses them in proofs and simplification.
6. Trigonometric equations finds every solution in a stated interval using graphs, symmetry and periodicity.

Two notational distinctions matter:

$$\sin^2x=(\sin x)^2,$$

but

$$\sin^{-1}x=\arcsin x,$$

which is the inverse sine function, not $1/\sin x$. The reciprocal of sine is $\csc x$.

Worked example 3: solve over a complete interval

Solve

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

for $0\leq x<2\pi$.

Treat the equation as a quadratic in $\cos x$:

$$(2\cos x-1)(\cos x-1)=0.$$

Therefore

$$\cos x=\frac12 \qquad\text{or}\qquad \cos x=1.$$

On $0\leq x<2\pi$,

$$\cos x=\frac12 \quad\Longrightarrow\quad x=\frac\pi3,\frac{5\pi}3,$$

and

$$\cos x=1 \quad\Longrightarrow\quad x=0.$$

Hence

$$\boxed{x=0,\frac\pi3,\frac{5\pi}3}.$$

The endpoint $2\pi$ is excluded. Although it gives the same cosine value as $0$, it must not be included in a half open interval.

Finding every solution

First find a reference angle, then use the signs in the relevant quadrants, then add or subtract whole periods. Finally check every answer against the stated interval. A calculator’s inverse function supplies a principal value, not automatically the complete solution set.

3. Non right angled triangles

7. Triangles and trigonometric rules covers the sine rule, cosine rule, triangle area formula and the ambiguous case.

For a triangle with side $a$ opposite angle $A$, and similarly for $b,B$ and $c,C$,

$$\boxed{\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}},$$ $$\boxed{a^2=b^2+c^2-2bc\cos A},$$

and

$$\boxed{\text{area}=\frac12bc\sin A}.$$

Choose a formula by matching the information, not by preference. The sine rule needs a known opposite side and angle pair. The cosine rule is suited to three sides, or two sides and their included angle.

Worked example 4: recognise the ambiguous case

In triangle $ABC$,

$$a=8, \qquad b=11, \qquad A=35^\circ.$$

Find the possible values of $B$.

The sine rule gives

$$\frac{\sin B}{11}=\frac{\sin35^\circ}{8},$$

so

$$\sin B=\frac{11\sin35^\circ}{8}\approx0.7887.$$

The calculator gives the acute possibility

$$B\approx\sin^{-1}(0.7887)\approx52.1^\circ.$$

Sine is also positive in the second quadrant, so

$$B\approx180^\circ-52.1^\circ=127.9^\circ.$$

Both are valid because

$$35^\circ+52.1^\circ<180^\circ$$

and

$$35^\circ+127.9^\circ<180^\circ.$$

Thus

$$\boxed{B\approx52.1^\circ\text{ or }127.9^\circ}.$$

When two sides and a non included angle are given, a second triangle may exist. Always test the supplementary angle rather than assuming the inverse sine output is unique.

4. Advanced identities and representations

Complete the pathway with the formulae and techniques that dominate the later A level course.

8. Compound angle and double angle formulae derives formulae for $\sin(A\pm B)$, $\cos(A\pm B)$ and $\tan(A\pm B)$, then applies them to exact values, proofs and equations.
9. Harmonic form rewrites $a\cos x+b\sin x$ as one shifted sine or cosine, exposing its amplitude, phase and range.
10. Small angle approximations explains why $\sin x\approx x$, $\tan x\approx x$ and $\cos x\approx1-\frac{x^2}{2}$ for small $x$ measured in radians.
11. Trigonometric proof and modelling combines identities, equations and functions in unfamiliar proofs and real contexts.

For example, harmonic form follows by expanding

$$R\cos(x-\alpha) =R\cos x\cos\alpha+R\sin x\sin\alpha.$$

Matching coefficients with $a\cos x+b\sin x$ gives

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

and hence

$$\boxed{R=\sqrt{a^2+b^2}}.$$

Worked example 5: use harmonic form to find a range

Find the maximum and minimum values of

$$5\cos x+12\sin x.$$

Write the expression as $R\cos(x-\alpha)$. Its amplitude is

$$R=\sqrt{5^2+12^2}=13.$$

Therefore

$$5\cos x+12\sin x=13\cos(x-\alpha)$$

for an appropriate $\alpha$. Since

$$-1\leq\cos(x-\alpha)\leq1,$$

it follows that

$$\boxed{-13\leq5\cos x+12\sin x\leq13}.$$

The exact phase angle is unnecessary when only the range is required. Recognising what the question needs can save substantial calculation.

Readiness check

Try these without notes. Each answer points to the lesson that repairs the underlying gap.

1. Exact values and quadrants

Find the exact value of

$$\sin150^\circ+\cos135^\circ.$$

Answer

Using reference angles and signs,

$$\sin150^\circ=\frac12, \qquad \cos135^\circ=-\frac{\sqrt2}{2}.$$

Therefore

$$\boxed{\sin150^\circ+\cos135^\circ=\frac{1-\sqrt2}{2}}.$$

Review exact trigonometric values if either value required a decimal.

2. Radians

Convert $225^\circ$ to radians.

Answer

Multiply by $\pi/180$:

$$225^\circ\times\frac{\pi}{180^\circ} =\frac{5\pi}{4}.$$

So the angle is $\boxed{\frac{5\pi}{4}}$ radians. Continue with radians, arcs and sectors.

3. Identities

Simplify

$$\frac{1-\cos^2x}{\sin x},$$

stating any restriction.

Answer

Since $1-\cos^2x=\sin^2x$,

$$\frac{1-\cos^2x}{\sin x} =\frac{\sin^2x}{\sin x} =\boxed{\sin x},$$

provided $\sin x\ne0$. The original expression is undefined when $x=n\pi$ for integers $n$. Study trigonometric identities if the restriction was easy to overlook.

4. Equations

Solve

$$\sin x=-\frac{\sqrt3}{2}$$

for $0\leq x<2\pi$.

Answer

The reference angle is $\pi/3$. Sine is negative in the third and fourth quadrants, so

$$\boxed{x=\frac{4\pi}{3},\frac{5\pi}{3}}.$$

Review trigonometric equations if you found only one value.

5. Proof or equation?

Is

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

an identity or an equation? Solve it for $0\leq x<2\pi$.

Answer

It is an equation because it is true only for particular values of $x$. Using $2\sin x\cos x=\sin2x$ gives

$$\sin2x=1.$$

As $0\leq x<2\pi$, the doubled angle satisfies $0\leq2x<4\pi$. Thus

$$2x=\frac\pi2,\frac{5\pi}{2},$$

so

$$\boxed{x=\frac\pi4,\frac{5\pi}{4}}.$$

Continue to compound angle and double angle formulae when basic equation solving is secure.

Common misconceptions

• A calculator mode is part of the mathematics. If an angle is written with a degree symbol, use degrees. If it contains $\pi$ or occurs in calculus, radians are normally intended.
• Inverse functions do not give every solution. They return a principal value. Use the graph, quadrants and period to complete the solution set.
• $\sin(A+B)$ is not $\sin A+\sin B$. Trigonometric functions are not distributive. Use the compound angle formula.
• Cancelling can change the domain. Dividing by $\sin x$ may discard solutions where $\sin x=0$. Record excluded values or rearrange without division.
• An identity is not proved by examples. Checking several angles can expose an error, but a proof requires valid algebra for all permitted values.
• Small angle approximations require radians. For example, $\sin x\approx x$ is false when the numerical value of $x$ is interpreted in degrees.

A reliable problem solving routine

1. Identify the angle unit and interval.
2. Sketch the relevant graph, circle or triangle if signs or solutions are unclear.
3. Keep exact values until a decimal accuracy is requested.
4. Transform the expression into a familiar identity, quadratic or single trigonometric function.
5. Find all possible angles using symmetry and periodicity.
6. Check domain restrictions, interval endpoints and the original equation.
7. In a model, interpret the answer with units and reject values that the context excludes.

Trigonometry becomes manageable when formulas are connected to the unit circle, graphs and algebra. Memorisation still matters, but structure tells you which formula to use and provides a way to check whether the result is plausible.