Trigonometry

The three graphs, and what each one does

The sine, cosine and tangent functions are defined for angles of any size, and the first thing to fix in your memory is the shape of each graph between $0^{\circ }$ and ${360}^{\circ }$. The graph of $y=\sin x$ starts at the origin, rises to a maximum of 1 at ${90}^{\circ }$, comes back through zero at ${180}^{\circ }$, falls to a minimum of $-1$ at ${270}^{\circ }$ and returns to zero at ${360}^{\circ }$. The graph of $y=\cos x$ is the same wave shape, but it starts at its highest point: $\cos 0^{\circ }=1$, it crosses the axis at ${90}^{\circ }$, reaches $-1$ at ${180}^{\circ }$, crosses again at ${270}^{\circ }$ and is back at 1 at ${360}^{\circ }$. Both graphs repeat every ${360}^{\circ }$, so everything you can see in one period happens again in every later period.

The graph of $y=\tan x$ is a different animal altogether. Since $\tan x=\frac{\sin x}{\cos x}$, the tangent is undefined wherever $\cos x=0$, and those values of $x$ give the vertical asymptotes at ${90}^{\circ }$, ${270}^{\circ }$ and every ${180}^{\circ }$ either side of them. Between two asymptotes the tangent graph rises from very large negative values, passes through zero halfway between them, and climbs to very large positive values. It repeats every ${180}^{\circ }$, and it has no maximum and no minimum at all.

Cambridge allows either degrees or radians on the horizontal axis, and you have to be fluent in both. In radians the same features sit at different numbers: $y=\sin x$ is a maximum at $\frac{\pi }{2}$, crosses zero at $\pi$ and $2\pi$, and is a minimum at $\frac{3\pi }{2}$; $y=\tan x$ has asymptotes at $\frac{\pi }{2}$ and $\frac{3\pi }{2}$. The interval a question gives you is the signal for which measure to use: $0^{\circ }\le x\le {360}^{\circ }$ expects degrees, and $0\le x\le 2\pi$ expects radians.

• $y=\sin x$: period ${360}^{\circ }$ (or $2\pi$), maximum 1 at ${90}^{\circ }$, minimum $-1$ at ${270}^{\circ }$
• $y=\cos x$: period ${360}^{\circ }$, maximum 1 at $0^{\circ }$ and at ${360}^{\circ }$, minimum $-1$ at ${180}^{\circ }$
• $y=\tan x$: period ${180}^{\circ }$ (or $\pi$), asymptotes at ${90}^{\circ }$, ${270}^{\circ }$, ${450}^{\circ }$, and every ${180}^{\circ }$ in between

Transformed graphs: $y=3\sin x$, $y=1-\cos 2x$ and $y=4\tan \frac{x}{2}+1$

A graph question usually asks you to sketch or describe a curve built out of one of the three, and every feature you need is readable from the numbers in the formula. For a curve of the form $y=A\sin (Bx)+C$, the amplitude is $|A|$, so the curve reaches $C+A$ and $C-A$; the period is $\frac{{360}^{\circ }}{B}$, or $\frac{2\pi }{B}$ in radians; and the $+C$ slides the whole graph up or down without changing its shape.

So $y=3\sin x$ is the ordinary sine curve stretched vertically by a factor of 3. It still has period ${360}^{\circ }$ and still crosses the axis at $0^{\circ }$, ${180}^{\circ }$ and ${360}^{\circ }$, but its maximum is 3 at ${90}^{\circ }$ and its minimum is $-3$ at ${270}^{\circ }$. The curve $y=1-\cos 2x$ takes three changes at once: the 2 halves the period to ${180}^{\circ }$, the minus sign turns the wave upside down so that it is lowest where $\cos 2x=1$, and the 1 lifts the whole graph so that it runs between 0 and 2. Its maximum is 2, and it is reached where $\cos 2x=-1$, that is at $x={90}^{\circ }$ and $x={270}^{\circ }$ in the first ${360}^{\circ }$.

For $y=4\tan \frac{x}{2}+1$ the halving of the angle doubles the period to ${360}^{\circ }$, so there is only one asymptote in the interval $0^{\circ }\le x\le {360}^{\circ }$, at $x={180}^{\circ }$, where $\frac{x}{2}={90}^{\circ }$. The curve passes through $(0^{\circ },1)$, because $\tan 0=0$, and the $+1$ raises every point of the tangent graph by one unit. There is still no maximum and no minimum.

Exact values from two triangles

Two right-angled triangles give you every exact value the syllabus asks for. Half of a square of side 1 is a ${45}^{\circ }$, ${45}^{\circ }$, ${90}^{\circ }$ triangle with two sides of 1 and a hypotenuse of $\sqrt{2}$. Half of an equilateral triangle of side 2 is a ${30}^{\circ }$, ${60}^{\circ }$, ${90}^{\circ }$ triangle with sides 1 and $\sqrt{3}$ and a hypotenuse of 2. Reading the three ratios off those triangles gives $\sin {45}^{\circ }=\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$, $\tan {45}^{\circ }=1$, $\sin {30}^{\circ }=\cos {60}^{\circ }=\frac{1}{2}$, $\cos {30}^{\circ }=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$, $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$ and $\tan {60}^{\circ }=\sqrt{3}$.

The values for the rest of the circle come from the symmetries of the graphs. An angle in the second quadrant is ${180}^{\circ }-\theta$, an angle in the third is ${180}^{\circ }+\theta$, and an angle in the fourth is ${360}^{\circ }-\theta$, where $\theta$ is the acute angle measured from the horizontal axis. The sine is positive in the first two quadrants, the cosine in the first and fourth, and the tangent in the first and third. Cambridge's own example is $\cos {150}^{\circ }=\cos ({180}^{\circ }-{30}^{\circ })=-\cos {30}^{\circ }=-\frac{\sqrt{3}}{2}$.

The reliable habit is to write the acute angle first, decide the sign from the quadrant, and only then quote the exact value. $\tan {300}^{\circ }=\tan ({360}^{\circ }-{60}^{\circ })=-\tan {60}^{\circ }=-\sqrt{3}$, and $\sin {210}^{\circ }=\sin ({180}^{\circ }+{30}^{\circ })=-\sin {30}^{\circ }=-\frac{1}{2}$. An answer given as a decimal such as $-0.866$ is not an exact value and will not score the mark, however close it is.

The two identities, and how to use them

Two identities do almost all the work in this topic: ${\sin }^2\theta +{\cos }^2\theta \equiv 1$ and $\frac{\sin \theta }{\cos \theta }\equiv \tan \theta$. The first is Pythagoras applied to a point on the unit circle, where the coordinates are $(\cos \theta ,\sin \theta )$; the second is the definition of the tangent. Both are true for every value of $\theta$ for which both sides exist, and both rearrange usefully: ${\sin }^2\theta =1-{\cos }^2\theta$, ${\cos }^2\theta =1-{\sin }^2\theta$ and $\sin \theta =\cos \theta \tan \theta$.

Read the notation carefully. ${\sin }^2\theta$ is shorthand for $(\sin \theta {)}^2$, the square of the sine of $\theta$. It is not $\sin ({\theta }^2)$, and it is not the same kind of object as ${\sin }^{-1}\theta$, which is an inverse function rather than a power. Mixing those two up is one of the most expensive mistakes in the paper.

To prove an identity, start from the more complicated side and change it, one step at a time, until it looks like the other side. Replace $1-{\cos }^2\theta$ with ${\sin }^2\theta$; split a $\tan \theta$ into $\frac{\sin \theta }{\cos \theta }$; cancel a common factor. Write down the identity you are quoting, because a scheme point is usually reserved for it, and never work on both sides at once: a proof that moves terms from one side to the other has assumed what it is trying to show.

The same two identities turn equations into ones you can solve. $2{\sin }^{2}x+\sin x-1=0$ is already a quadratic in $\sin x$ and factorises as $(2\sin x-1)(\sin x+1)=0$. An equation such as $3{\cos }^{2}x=1+\sin x$ is not, until you replace ${\cos }^{2}x$ with $1-{\sin }^{2}x$ and collect the terms: $3(1-{\sin }^{2}x)=1+\sin x$ becomes $3{\sin }^{2}x+\sin x-2=0$.

• ${\sin }^2\theta +{\cos }^2\theta =1$ gives both ${\sin }^2\theta =1-{\cos }^2\theta$ and ${\cos }^2\theta =1-{\sin }^2\theta$
• $\tan \theta =\frac{\sin \theta }{\cos \theta }$, so $\tan \theta \cos \theta =\sin \theta$
• Proving an identity: work on one side only, quote the identity you use, and finish by saying that the two sides are equal
• Solving an equation with ${\cos }^2$ and $\sin$ together: replace one of them with $1$ minus the other, then factorise

${\sin }^{-1}x$, ${\cos }^{-1}x$ and ${\tan }^{-1}x$: the principal values

The notation ${\sin }^{-1}x$ means the angle whose sine is $x$. Because the sine graph takes the same value twice in every ${360}^{\circ }$, there are infinitely many such angles, so the notation is defined to return exactly one of them, called the principal value. The principal value of ${\sin }^{-1}x$ lies between $-{90}^{\circ }$ and ${90}^{\circ }$; the principal value of ${\cos }^{-1}x$ lies between $0^{\circ }$ and ${180}^{\circ }$; the principal value of ${\tan }^{-1}x$ lies strictly between $-{90}^{\circ }$ and ${90}^{\circ }$. These three ranges are printed in the formula booklet, so you do not have to memorise them, but you do have to know what they mean.

${\sin }^{-1}\left(\frac{1}{2}\right)={30}^{\circ }$ and not ${150}^{\circ }$, because ${150}^{\circ }$ is outside the range. ${\cos }^{-1}(-\frac{1}{2})={120}^{\circ }$, because the cosine inverse is allowed to give an obtuse angle. ${\tan }^{-1}(-1)=-{45}^{\circ }$ and not ${135}^{\circ }$, because the tangent inverse returns a negative angle rather than an angle in the second quadrant.

The three are inverse functions in the same sense as $f^{-1}$, so ${\sin }^{-1}(\sin x)=x$ only when $x$ is inside the principal range. ${\sin }^{-1}(\sin {200}^{\circ })$ is not ${200}^{\circ }$: ${200}^{\circ }$ lies outside the range, and $\sin {200}^{\circ }=-\sin {20}^{\circ }$, so the principal value is $-{20}^{\circ }$. Questions of exactly that shape appear as one-mark starters, and the mark is for noticing the range rather than for pressing a button on the calculator.

Solving a trigonometrical equation in a stated interval

Every equation of this kind follows the same four steps. Rearrange until a single function of a single angle stands alone on the left-hand side. Find the principal value with the appropriate inverse function. Use the graph, or the symmetry of the quadrant rules, to write down every angle in the given interval that has that value. Check each answer against the interval before you stop writing.

How many solutions to expect follows from the periods. In an interval of length ${360}^{\circ }$, the equations $\sin x=k$ and $\cos x=k$ each have two solutions when $-1<k<1$, while $\tan x=k$ has one solution in every ${180}^{\circ }$, so it has two in ${360}^{\circ }$. Double the interval and the count doubles: $\sin x=k$ has four solutions in $0^{\circ }\le x\le {720}^{\circ }$.

Take $\sin x=-\frac{1}{2}$ in $0^{\circ }\le x\le {360}^{\circ }$. The principal value is ${\sin }^{-1}(-\frac{1}{2})=-{30}^{\circ }$, which is outside the interval, so add ${360}^{\circ }$ to bring it in: ${330}^{\circ }$. The second solution comes from the other half of the sine wave, at ${180}^{\circ }+{30}^{\circ }={210}^{\circ }$. Both lie in the interval, so the answers are ${210}^{\circ }$ and ${330}^{\circ }$, and a sketch showing the line $y=-\frac{1}{2}$ cutting the curve twice is the quickest check that you have not missed one.

When the angle in the equation is doubled or halved, change the interval to match before you start. For $\cos 2x=\frac{1}{2}$ with $0\le x\le 2\pi$ you are really solving for $2x$ in $0\le 2x\le 4\pi$, and there are four solutions in that wider interval: $2x=\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}$. Halving each gives $x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$. Forgetting to widen the interval is the single commonest way to lose the last two marks of a question like this.

If the equation is a quadratic in $\sin x$ or $\cos x$, factorise it, solve the two linear equations that come out, and then find the angles for each value separately. Keep the two sets of angles apart until the final line, and count them: $\sin x=\frac{1}{2}$ gives two angles in ${360}^{\circ }$, but $\sin x=-1$ gives only one.

• $\sin x=k$: two solutions per ${360}^{\circ }$, at $\alpha$ and ${180}^{\circ }-\alpha$ when $\alpha$ is the acute angle
• $\cos x=k$: two solutions per ${360}^{\circ }$, at $\alpha$ and ${360}^{\circ }-\alpha$
• $\tan x=k$: one solution per ${180}^{\circ }$, at $\alpha$ and $\alpha +{180}^{\circ }$
• Angle doubled in the equation, interval doubled for the working: solve for $2x$ first, then halve every answer