| Tool | Identities |
|---|---|
| Definitions | $\tan x = \dfrac{\sin x}{\cos x}$, $\sec x = \dfrac{1}{\cos x}$, $\operatorname{cosec} x = \dfrac{1}{\sin x}$, $\cot x = \dfrac{\cos x}{\sin x}$ |
| Pythagorean | $\sin^2 x + \cos^2 x \equiv 1$, $1 + \tan^2 x \equiv \sec^2 x$, $1 + \cot^2 x \equiv \operatorname{cosec}^2 x$ |
| Compound angle | $\sin(A \pm B)$, $\cos(A \pm B)$, $\tan(A \pm B)$ |
| Double angle | $\sin 2x = 2\sin x\cos x$, $\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1$ |
- If you cannot see a route, rewrite everything in terms of $\sin$ and $\cos$.
- Combine fractions over a common denominator.
- Look for $1 - \sin^2 x$, $1 - \cos^2 x$, $1 \pm \cos 2x$ — they simplify.
- Factorise and cancel. Keep the target in view: if the answer has $\cot x$, you want $\cos x$ over $\sin x$.
Prove that $\dfrac{\cos x}{1 - \sin x} + \dfrac{\cos x}{1 + \sin x} \equiv 2\sec x$.
Prove that $\operatorname{cosec} 2x + \cot 2x \equiv \cot x$.
Prove that $\dfrac{\sin 3x}{\sin x} - \dfrac{\cos 3x}{\cos x} \equiv 2$.
The method is the same as in Year 1; what changes is the toolkit for getting to a single function, and the interval is usually in radians.
Solve $2\sin 2\theta = 3\tan\theta$ for $0 \leqslant \theta \leqslant 2\pi$.
Solve $2\cos^2 x + 3\sin x = 3$ for $0 \leqslant x \leqslant 2\pi$.
Anything that repeats regularly — tides, a Ferris wheel, hours of daylight, a vibrating spring — can be modelled by a sine or cosine function.
centre line $A$ · amplitude $|B|$ · maximum $A + |B|$, minimum $A - |B|$ · period $\dfrac{2\pi}{\omega}$
The depth of water in a harbour is $D = 5 + 3\cos\left(\dfrac{\pi t}{6}\right)$ metres, $t$ hours after midnight. A ship needs at least $6.5$ m of water. For how long in each tidal cycle can it use the harbour?
A Ferris wheel has radius $20$ m and its centre is $22$ m above the ground. It turns once every $10$ minutes. A passenger boards at the lowest point at $t = 0$. (a) Find a model for the height $h$ m after $t$ minutes. (b) Find the height after $2$ minutes. (c) For how long in each revolution is the passenger more than $30$ m above the ground?
The number of hours of daylight in a UK city on day $t$ of the year is modelled by $H = 12 + 4.5\sin\left(\dfrac{2\pi(t - 80)}{365}\right)$. (a) Find the greatest number of hours of daylight and the day it occurs. (b) For how many days is there more than $15$ hours of daylight? (c) Give one limitation of the model.
Proof layout
Start from one side; end on the other. Finish with $\blacksquare$.
Stuck?
Convert to $\sin$ and $\cos$; combine fractions.
$1 \pm \cos 2x$
$1 + \cos 2x = 2\cos^2 x$; $1 - \cos 2x = 2\sin^2 x$
Equations
Factorise; keep both $\pm$ roots.
Period
$\dfrac{2\pi}{\omega}$ for $\sin(\omega t + c)$.
Max / min
$A \pm |B|$ for $A + B\sin(\ldots)$.
Start at the bottom
Use $A - B\cos(\omega t)$.
Duration
Solve the inequality for one full cycle.
Prove that $(\sin x + \cos x)^2 \equiv 1 + \sin 2x$.
▶ Show solution
LHS $= \sin^2 x + 2\sin x\cos x + \cos^2 x = 1 + \sin 2x = $ RHS $\blacksquare$
Prove that $\dfrac{1 - \cos 2x}{1 + \cos 2x} \equiv \tan^2 x$.
▶ Show solution
LHS $= \dfrac{2\sin^2 x}{2\cos^2 x} = \tan^2 x = $ RHS $\blacksquare$
Prove that $\cot x - \tan x \equiv 2\cot 2x$.
▶ Show solution
LHS $= \dfrac{\cos x}{\sin x} - \dfrac{\sin x}{\cos x} = \dfrac{\cos^2 x - \sin^2 x}{\sin x\cos x} = \dfrac{\cos 2x}{\frac12\sin 2x} = 2\cot 2x = $ RHS $\blacksquare$
Prove that $\tan x + \cot x \equiv 2\operatorname{cosec} 2x$.
▶ Show solution
LHS $= \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x} = \dfrac{2}{\sin 2x} = 2\operatorname{cosec} 2x$ $\blacksquare$
Solve $\sin 2x = \cos x$ for $0 \leqslant x \leqslant 2\pi$.
▶ Show solution
$2\sin x\cos x - \cos x = 0 \Rightarrow \cos x(2\sin x - 1) = 0$.
$\cos x = 0$: $x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$. $\sin x = \tfrac12$: $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$.
Solve $\tan 2x = 3\tan x$ for $0 \leqslant x \leqslant \pi$.
▶ Show solution
Let $t = \tan x$: $\dfrac{2t}{1 - t^2} = 3t \Rightarrow 2t = 3t - 3t^3 \Rightarrow t(3t^2 - 1) = 0$.
$t = 0$: $x = 0, \pi$. $t = \pm\tfrac{1}{\sqrt3}$: $x = \tfrac{\pi}{6}, \tfrac{5\pi}{6}$.
(None of these makes $1 - t^2 = 0$, so all are valid.)
For $y = 3 + 2\sin\left(4t - \tfrac{\pi}{3}\right)$, state the amplitude, period, maximum value, and the smallest positive $t$ at which the maximum occurs.
▶ Show solution
Amplitude $2$; period $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$; maximum $3 + 2 = 5$.
$4t - \tfrac{\pi}{3} = \tfrac{\pi}{2} \Rightarrow 4t = \tfrac{5\pi}{6} \Rightarrow t = \tfrac{5\pi}{24}$.
The displacement of a mass on a spring is $x = 4\cos 2t + 3\sin 2t$ cm, $t$ seconds after release. Find the amplitude of the motion and the first time at which $x = 0$.
▶ Show solution
$x = R\cos(2t - \alpha)$ with $R\cos\alpha = 4$, $R\sin\alpha = 3$: $R = 5$ cm, $\alpha = \arctan\tfrac34 = 0.6435$.
$x = 0$ first when $2t - 0.6435 = \tfrac{\pi}{2}$, so $t = \dfrac{1.5708 + 0.6435}{2} = 1.11$ s (3 s.f.).
A Ferris wheel of radius $15$ m has its centre $17$ m above the ground and turns once every $8$ minutes. A passenger boards at the bottom at $t = 0$. Write down a model for the height $h$ m after $t$ minutes, and find the first time the passenger is $25$ m above the ground.
▶ Show solution
$\omega = \tfrac{2\pi}{8} = \tfrac{\pi}{4}$, and starting at the bottom: $h = 17 - 15\cos\left(\tfrac{\pi t}{4}\right)$.
$17 - 15\cos\left(\tfrac{\pi t}{4}\right) = 25 \Rightarrow \cos\left(\tfrac{\pi t}{4}\right) = -\tfrac{8}{15}$.
$\tfrac{\pi t}{4} = \arccos\left(-\tfrac{8}{15}\right) = 2.1333$, so $t = \dfrac{4 \times 2.1333}{\pi} = 2.72$ minutes (3 s.f.).
The depth of water at a harbour entrance is modelled by $D = 7 + 2.5\sin\left(\dfrac{\pi t}{6.2}\right)$ metres, where $t$ is the time in hours after midnight.
(a) State the period of the model and explain why it is realistic for tides. (b) Find the maximum and minimum depths and the first times at which they occur. (c) A boat needs a depth of more than $8$ m. Find the time window during the first tidal cycle when it can enter, to the nearest minute. (d) Suggest one way in which the model could be refined.
▶ Show solution
(a) Period $= \dfrac{2\pi}{\pi/6.2} = 12.4$ hours. Real tides have two highs and two lows a day, about $12$ hours $25$ minutes apart, so this matches well.
(b) Maximum $9.5$ m when $\tfrac{\pi t}{6.2} = \tfrac{\pi}{2}$, i.e. $t = 3.1$ h (03:06). Minimum $4.5$ m when $\tfrac{\pi t}{6.2} = \tfrac{3\pi}{2}$, i.e. $t = 9.3$ h (09:18).
(c) $2.5\sin\left(\tfrac{\pi t}{6.2}\right) > 1 \Rightarrow \sin\left(\tfrac{\pi t}{6.2}\right) > 0.4$. $\arcsin 0.4 = 0.4115$.
$0.4115 < \tfrac{\pi t}{6.2} < \pi - 0.4115 = 2.7301$, so $0.812 < t < 5.388$.
The boat can enter between 00:49 and 05:23 — a window of about $4$ hours $35$ minutes.
(d) The heights of high and low tide vary over a lunar month (spring and neap tides), so the amplitude $2.5$ could itself be made to vary with time; or the model could be fitted to real tide-table data.