Trigonometry

Secant, cosecant and cotangent Paper 3

Three more trigonometric functions are defined as reciprocals:
$$\sec \theta =\frac{1}{\cos \theta }\mathrm{cosec}\theta =\frac{1}{\sin \theta }\cot \theta =\frac{1}{\tan \theta }=\frac{\cos \theta }{\sin \theta }$$
Each is not defined where its denominator is zero: $\sec \theta$ where $\cos \theta =0$ (at $\pm {90}^{°},\pm {270}^{°},\dots$), and $\mathrm{cosec}\theta$ and $\cot \theta$ where $\sin \theta =0$ (at $0^{°},\pm {180}^{°},\dots$). At those angles the graphs have vertical asymptotes.

The properties follow from the parent functions, for angles of any size:

• $\sec \theta$ and $\mathrm{cosec}\theta$ have period ${360}^{°}$ ($2\pi$). Since $\left|\cos \theta \right|⩽1$, its reciprocal satisfies $\left|\sec \theta \right|⩾1$: the range is $\sec \theta ⩽-1$ or $\sec \theta ⩾1$, and the same for $\mathrm{cosec}\theta$.
• $\cot \theta$ has period ${180}^{°}$ ($\pi$) and takes every real value. $\cot \theta =0$ where $\cos \theta =0$.
• Each has the same sign as its parent: $\sec \theta$ is negative exactly where $\cos \theta$ is, in the second and third quadrants.
• Where the parent has a maximum or minimum of $\pm 1$, the reciprocal has a minimum or maximum of $\pm 1$ at the same angle.

Exact values come from the familiar ones: $\sec {60}^{°}=\frac{1}{\frac{1}{2}}=2$, $\mathrm{cosec}\frac{\pi }{4}=\sqrt{2}$, $\cot {150}^{°}=-\sqrt{3}$. Calculators have no sec, cosec or cot keys, so always convert: to solve $\sec x=3$, solve $\cos x=\frac{1}{3}$.

$\sec x$ touches $\cos x$ where $\cos x=\pm 1$ and has an asymptote wherever $\cos x=0$.

$\mathrm{cosec}x$ and $\cot x$ both have asymptotes where $\sin x=0$; $\cot x$ has period ${180}^{°}$.

Identities from Pythagoras Paper 3

Dividing ${\sin }^2\theta +{\cos }^2\theta \equiv 1$ by ${\cos }^2\theta$, and then by ${\sin }^2\theta$, gives two more identities:
$${\sec }^2\theta \equiv 1+{\tan }^2\theta {\mathrm{cosec}}^2\theta \equiv 1+{\cot }^2\theta$$
Both are in MF19. Their main use is in equations that mix two functions: replace ${\sec }^2\theta$ by $1+{\tan }^2\theta$ to get a quadratic in $\tan \theta$ alone, or ${\mathrm{cosec}}^2\theta$ by $1+{\cot }^2\theta$ to get one in $\cot \theta$. Then factorise, and solve each factor over the whole interval, using the CAST diagram or a graph to find every angle.

Compound angle formulae Paper 3

The sine, cosine and tangent of a sum or difference of angles are given in MF19:
$$\sin (A\pm B)\equiv \sin A\cos B\pm \cos A\sin B$$
$$\cos (A\pm B)\equiv \cos A\cos B\mp \sin A\sin B$$
$$\tan (A\pm B)\equiv \frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$$
Watch the signs: in the cosine formula the sign on the right is the opposite of the one on the left. And $\sin (A+B)$ is not $\sin A+\sin B$: try $A=B={90}^{°}$.

Uses:

• Exact values: ${15}^{°}={45}^{°}-{30}^{°}$, ${75}^{°}={45}^{°}+{30}^{°}$, ${105}^{°}={60}^{°}+{45}^{°}$.
• Simplifying an expression such as $\cos (x-{30}^{°})-3\sin (x-{60}^{°})$ to a form $p\sin x+q\cos x$, by expanding each bracket and collecting terms.
• Solving an equation such as $\sin (x+{30}^{°})=2\cos x$: expand, collect, and divide by $\cos x$ to reach $\tan x=k$.
• Finding $\sin (A+B)$ when only $\sin A$ and $\cos B$ are known: draw right-angled triangles, or use ${\sin }^2+{\cos }^2=1$, to find the other ratios, with signs from the quadrant.

Double angle formulae Paper 3

Put $B=A$ in the compound angle formulae:
$$\sin 2A\equiv 2\sin A\cos A\tan 2A\equiv \frac{2\tan A}{1-{\tan }^{2}A}$$
$$\cos 2A\equiv {\cos }^{2}A-{\sin }^{2}A\equiv 2{\cos }^{2}A-1\equiv 1-2{\sin }^{2}A$$
The three forms of $\cos 2A$ come from replacing ${\sin }^{2}A$ or ${\cos }^{2}A$ using ${\sin }^{2}A+{\cos }^{2}A=1$. Choose the form that matches the rest of the equation: with $\sin x$ use $1-2{\sin }^{2}x$; with $\cos x$ use $2{\cos }^{2}x-1$. The result is a quadratic in one function.

Rearranged, the last two give ${\cos }^{2}A=\frac{1}{2}(1+\cos 2A)$ and ${\sin }^{2}A=\frac{1}{2}(1-\cos 2A)$, which turn squares into first powers.

With $\sin 2x$, write $2\sin x\cos x$ and factorise; never divide by $\sin x$ or $\cos x$, because that loses the solutions where it is zero.

Expressing a sin θ + b cos θ as a single function Paper 3

Any expression $a\sin \theta +b\cos \theta$ is a single sine wave in disguise. Expanding the target form,
$$R\sin (\theta +\alpha )\equiv R\sin \theta \cos \alpha +R\cos \theta \sin \alpha$$
and matching coefficients gives $R\cos \alpha =a$ and $R\sin \alpha =b$. Squaring and adding, and dividing, gives
$$R=\sqrt{a^2+b^2}\tan \alpha =\frac{b}{a}$$
Take $R>0$, and use the given form to decide the signs: in $R\sin (\theta -\alpha )$ the expansion is $R\sin \theta \cos \alpha -R\cos \theta \sin \alpha$, so $5\sin \theta -2\cos \theta$ gives $R\cos \alpha =5$ and $R\sin \alpha =2$, both positive, and $\alpha$ is acute. The same matching works for $R\cos (\theta \pm \alpha )$, which expands to $R\cos \theta \cos \alpha \mp R\sin \theta \sin \alpha$.

Once in this form:

• the greatest and least values are $R$ and $-R$, when the sine or cosine is 1 or $-1$;
• an equation $a\sin \theta +b\cos \theta =c$ becomes $\sin (\theta +\alpha )=\frac{c}{R}$. Adjust the interval for $\theta +\alpha$ before finding every angle, then subtract $\alpha$.

Keep $\alpha$ to more figures than the final answer needs, and give it to the accuracy asked for, often 2 decimal places in degrees.

Choosing an identity Paper 3

The skill examined most is choosing the identity that fits. Some guides:

• ${\sec }^2$ with $\tan$, or ${\mathrm{cosec}}^2$ with $\cot$: use the Pythagorean identities, as in $2{\sec }^2\theta -\tan \theta =5$.
• $\tan$ with $\cot$: write both in sine and cosine, or use $\cot \theta =\frac{1}{\tan \theta }$. For example $\tan \theta +\cot \theta =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{2}{\sin 2\theta }$, so $\tan \theta +\cot \theta =4$ reduces to $\sin 2\theta =\frac{1}{2}$.
• A double angle with a single angle: replace the double angle, choosing the $\cos 2A$ form that matches.
• $a\sin \theta +b\cos \theta =c$: use the $R$ form, as in $3\cos \theta +2\sin \theta =1$.
• A compound angle in a bracket: expand it.

To prove an identity, start from one side (usually the more complicated) and transform it step by step into the other, quoting each identity used. Do not treat it as an equation by working on both sides at once.

In equations, dividing by a trigonometric function, or squaring, changes the solution set. Factorise instead of dividing, and check any answers found after squaring.