Trigonometry

Angles of any size Paper 1

In GCSE work, $\sin \theta$, $\cos \theta$ and $\tan \theta$ are ratios of sides in a right-angled triangle, so they only make sense for acute angles. At A Level we need them for any angle, including obtuse, reflex and negative ones, so we use a circle instead.

Draw a circle of radius 1 centred at the origin. Start on the positive $x$-axis and turn anticlockwise through the angle $\theta$ (a clockwise turn is a negative angle). The point you reach on the circle has coordinates $(\cos \theta ,\sin \theta )$, and $\tan \theta =\frac{\sin \theta }{\cos \theta }$ is the gradient of the radius. For an acute angle this agrees with SOHCAHTOA, because the hypotenuse is 1.

The signs follow from which quadrant the point lands in:

• $0^{°}$ to ${90}^{°}$: all three are positive.
• ${90}^{°}$ to ${180}^{°}$: only $\sin$ is positive ($x<0$, $y>0$).
• ${180}^{°}$ to ${270}^{°}$: only $\tan$ is positive (both coordinates negative).
• ${270}^{°}$ to ${360}^{°}$: only $\cos$ is positive.

Many students remember this as CAST, read anticlockwise from the fourth quadrant. Each angle also has a basic acute angle, the angle its radius makes with the $x$-axis. For ${150}^{°}$ it is ${30}^{°}$; for ${225}^{°}$ it is ${45}^{°}$; for ${300}^{°}$ it is ${60}^{°}$. The size of the ratio comes from the basic angle and the sign comes from the quadrant. For example $\cos {150}^{°}=-\cos {30}^{°}$, because ${150}^{°}$ is in the second quadrant, where cosine is negative.

Radians. Pure Mathematics also measures angles in radians, where $\pi$ radians $={180}^{°}$. So $\frac{1}{6}\pi ={30}^{°}$, $\frac{1}{4}\pi ={45}^{°}$, $\frac{1}{3}\pi ={60}^{°}$, $\frac{1}{2}\pi ={90}^{°}$ and $2\pi ={360}^{°}$. Everything in this lesson works in either unit: an interval written with $\pi$, such as $0⩽x⩽2\pi$, tells you to work and answer in radians, and your calculator must be in the matching mode.

Graphs of sine, cosine and tangent Paper 1

As $\theta$ turns round the circle, the height of the point traces the graph of $y=\sin \theta$ and its horizontal position traces $y=\cos \theta$. Both waves repeat every ${360}^{°}$ (or $2\pi$): they are periodic with period ${360}^{°}$.

• $y=\sin x$ starts at $(0,0)$, reaches its maximum 1 at ${90}^{°}$, crosses the axis at ${180}^{°}$, reaches $-1$ at ${270}^{°}$ and returns to 0 at ${360}^{°}$.
• $y=\cos x$ is the same wave moved ${90}^{°}$ to the left: it starts at its maximum $(0,1)$, crosses at ${90}^{°}$ and ${270}^{°}$ and has its minimum $-1$ at ${180}^{°}$.
• $y=\tan x$ has period ${180}^{°}$. It passes through the origin, increases through every value, and has vertical asymptotes at $x={90}^{°}$, ${270}^{°}$, ..., where $\cos x=0$ and $\tan x$ is not defined.

The graphs carry the symmetry that solving equations depends on. The sine curve is symmetrical about $x={90}^{°}$, so $\sin ({180}^{°}-\theta )=\sin \theta$. The cosine curve is symmetrical about the $y$-axis and about $x={180}^{°}$, so $\cos ({360}^{°}-\theta )=\cos \theta$. The tangent curve repeats every ${180}^{°}$, so $\tan ({180}^{°}+\theta )=\tan \theta$.

One full cycle of each wave. The cosine curve is the sine curve translated ${90}^{°}$ to the left.

The tangent graph repeats every ${180}^{°}$ and has asymptotes where $\cos x=0$.

Transformations of the graphs Paper 1

Exam questions use the basic graphs with simple changes. Each change has a predictable effect, and you can always check it by plotting one or two points.

• $y=a\sin x$: a vertical stretch by factor $a$. The amplitude (the height of a peak above the centre line) becomes $a$, so $y=3\sin x$ runs from $-3$ to $3$.
• $y=\sin bx$: a horizontal stretch by factor $\frac{1}{b}$. The period becomes $\frac{{360}^{°}}{b}$, so $y=\cos 2x$ completes two cycles between $0^{°}$ and ${360}^{°}$.
• $y=c+\sin x$: a translation up by $c$. The centre line moves to $y=c$.
• $y=\sin (x+c)$: a translation of $c$ to the left (a negative $c$ moves it right). For example $y=\tan (x+\frac{1}{4}\pi )$ is $y=\tan x$ moved $\frac{1}{4}\pi$ to the left, so it crosses the $x$-axis at $x=-\frac{1}{4}\pi$ and $x=\frac{3}{4}\pi$, and its asymptotes move to $x=\frac{1}{4}\pi$ and $x=\frac{5}{4}\pi$.
• $y=-\cos x$: a reflection in the $x$-axis.

A graph such as $y=1-\cos 2x$ combines three of these: $\cos 2x$ has period ${180}^{°}$, the minus sign turns it upside down, and the 1 lifts it so that it runs from 0 (when $\cos 2x=1$) to 2 (when $\cos 2x=-1$).

The greatest and least values of $a+b\cos x$ (with $b>0$) are $a+b$ and $a-b$, because $\cos x$ itself only runs from $-1$ to $1$. That simple fact answers many questions about ranges and about how many solutions an equation has.

Two complete cycles between $0^{°}$ and ${360}^{°}$, between the lines $y=0$ and $y=2$.

Exact values Paper 1

Papers expect you to know the exact values for ${30}^{°}$, ${45}^{°}$ and ${60}^{°}$, and to use them for related angles in any quadrant. They come from two triangles, so you never need to memorise a table blindly.

• Half of a square with sides 1 is a right-angled isosceles triangle with angles ${45}^{°}$ and hypotenuse $\sqrt{2}$. So $\sin {45}^{°}=\cos {45}^{°}=\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2}$, and $\tan {45}^{°}=1$.
• Half of an equilateral triangle with sides 2 has sides 1, $\sqrt{3}$ and 2, and angles ${30}^{°}$, ${60}^{°}$ and ${90}^{°}$. So $\sin {30}^{°}=\frac{1}{2}$, $\cos {30}^{°}=\frac{1}{2}\sqrt{3}$, $\tan {30}^{°}=\frac{1}{\sqrt{3}}=\frac{1}{3}\sqrt{3}$, and $\sin {60}^{°}=\frac{1}{2}\sqrt{3}$, $\cos {60}^{°}=\frac{1}{2}$, $\tan {60}^{°}=\sqrt{3}$.

Also learn the values at the axes: $\sin 0=0$, $\cos 0=1$, $\sin {90}^{°}=1$, $\cos {90}^{°}=0$, $\cos {180}^{°}=-1$, and $\tan {90}^{°}$ is not defined.

For a related angle, find the basic acute angle and attach the sign from the quadrant. $\cos {150}^{°}$: the basic angle is ${30}^{°}$ and ${150}^{°}$ is in the second quadrant, where cosine is negative, so $\cos {150}^{°}=-\frac{1}{2}\sqrt{3}$. In radians, $\sin \frac{3}{4}\pi$: the basic angle is $\frac{1}{4}\pi$ and the angle is in the second quadrant, where sine is positive, so $\sin \frac{3}{4}\pi =\frac{1}{2}\sqrt{2}$.

When a question says "find the exact value", a decimal from the calculator earns no accuracy mark, even when it is correct to many figures.

Inverse trigonometric functions and principal values Paper 1

The equation $\sin x=\frac{1}{2}$ has infinitely many solutions: ${30}^{°}$, ${150}^{°}$, ${390}^{°}$, and so on. So "the angle whose sine is $\frac{1}{2}$" is not one angle unless we agree which one we mean. The agreed choice is the principal value, and it is what your calculator gives:

• ${\sin }^{-1}x$ gives an angle from $-{90}^{°}$ to ${90}^{°}$ (from $-\frac{1}{2}\pi$ to $\frac{1}{2}\pi$);
• ${\cos }^{-1}x$ gives an angle from $0^{°}$ to ${180}^{°}$ (from 0 to $\pi$);
• ${\tan }^{-1}x$ gives an angle strictly between $-{90}^{°}$ and ${90}^{°}$.

These ranges are not arbitrary. On each one the function takes every value it can take exactly once, so sine, cosine and tangent restricted to those intervals are one-one functions and have inverse functions. That is exactly the idea of an inverse function from the Functions topic: ${\sin }^{-1}$ undoes $\sin$, but only on the restricted domain. So ${\sin }^{-1}(\sin {150}^{°})={30}^{°}$, not ${150}^{°}$.

Note that ${\sin }^{-1}x$ does not mean $\frac{1}{\sin x}$. The $-1$ here means "inverse function", not a power.

For example, ${\cos }^{-1}(-\frac{1}{2})={120}^{°}$, because ${120}^{°}$ is the only angle from $0^{°}$ to ${180}^{°}$ with cosine $-\frac{1}{2}$, while ${\sin }^{-1}(-\frac{1}{2})=-{30}^{°}$, a negative angle, because sine's principal values include the fourth quadrant.

The two identities Paper 1

An identity is true for every value of the variable for which both sides are defined, and is written with $\equiv$. Two identities are needed at this stage, and both come straight from the unit circle:
$$\tan \theta \equiv \frac{\sin \theta }{\cos \theta }{\sin }^2\theta +{\cos }^2\theta \equiv 1$$
The first is the gradient of the radius, $\frac{y}{x}$. The second is Pythagoras' theorem for the right-angled triangle with hypotenuse 1 and legs $\cos \theta$ and $\sin \theta$. (${\sin }^2\theta$ means $(\sin \theta {)}^2$.)

They are used in three ways, exactly as the syllabus lists.

• Simplifying. For example $\frac{1-{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{{\sin }^2\theta }{\sin \theta \cos \theta }=\frac{\sin \theta }{\cos \theta }=\tan \theta$.
• Proving identities. Start from one side (usually the more complicated one) and work to the other using steps that are always true. Do not write the identity down and "do the same to both sides": that assumes what you are trying to prove. Combining fractions over a common denominator, factorising and replacing $1-{\cos }^2\theta$ by ${\sin }^2\theta$ are the usual moves.
• Solving equations. Replace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ (or the reverse) so that only one trigonometric function remains, and replace $\frac{\sin \theta }{\cos \theta }$ by $\tan \theta$.

The rearrangements ${\sin }^2\theta =1-{\cos }^2\theta$ and ${\cos }^2\theta =1-{\sin }^2\theta$ are the forms you will use most.

Solving trigonometric equations in an interval Paper 1

Questions ask for all the solutions in a stated interval. General forms of solution are not needed, but missing a solution in the interval costs a mark, so the method has to be systematic.

One function, one angle. For $\sin x=k$ (with $-1⩽k⩽1$):

1. Find the principal value $\alpha ={\sin }^{-1}k$ on your calculator.
2. Use the symmetry of the graph to find the second value in one cycle: ${180}^{°}-\alpha$ for sine, ${360}^{°}-\alpha$ for cosine, ${180}^{°}+\alpha$ for tangent.
3. Add or subtract ${360}^{°}$ (or ${180}^{°}$ for tangent) to find any others in the interval, and discard values outside it.

A quick sketch of the graph with the line $y=k$ shows how many solutions to expect, and is the best defence against losing one. If $k>1$ or $k<-1$ there are no solutions to $\sin x=k$ or $\cos x=k$.

A multiple or shifted angle. For $\sin 2x=k$ with $0^{°}⩽x⩽{180}^{°}$, let $u=2x$, so $0^{°}⩽u⩽{360}^{°}$. Find every $u$ in the new interval, then halve each one. The most common error is to find $u$ only in the original interval and lose half the solutions. For $\tan (x+{30}^{°})=k$, shift the interval: $u=x+{30}^{°}$ runs from ${30}^{°}$ to ${390}^{°}$ when $x$ runs from $0^{°}$ to ${360}^{°}$.

Sine and cosine of the same angle. $a\sin x=b\cos x$ becomes $\tan x=\frac{b}{a}$ after dividing by $\cos x$. That is safe here, because if $\cos x=0$ then $\sin x=\pm 1$ and the equation would be false.

A common factor. In $\sin x\cos x=\frac{1}{2}\cos x$ you must not divide by $\cos x$, because $\cos x=0$ does make this equation true. Factorise instead: $\cos x(\sin x-\frac{1}{2})=0$, so $\cos x=0$ or $\sin x=\frac{1}{2}$. Dividing throws the solutions of $\cos x=0$ away.

Quadratic in a trigonometric function. For $3{\sin }^2\theta -5\cos \theta -1=0$, use ${\sin }^2\theta =1-{\cos }^2\theta$ so that everything is in $\cos \theta$, rearrange into a quadratic, factorise (or use the formula), then solve each factor. Reject any value outside $-1$ to $1$, and say why.

Give angles in degrees to 1 decimal place and radians to 3 significant figures unless an exact answer is asked for, and keep full calculator accuracy until the final step.