The R cos(x + a) Form – A Level Maths Year 2

1 Two Waves Make One

Any expression $a\cos\theta + b\sin\theta$ is a single sine wave in disguise — stretched to amplitude $R$ and shifted by an angle $\alpha$. Writing it as $R\cos(\theta \pm \alpha)$ or $R\sin(\theta \pm \alpha)$ turns a hard equation with two trig functions into an easy one with one.

The four forms (with $R > 0$ and $\alpha$ acute)

$R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha$
$R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha$
$R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$
$R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$

Choosing the form. Match the signs. $3\sin\theta + 4\cos\theta$ (both $+$, $\sin$ first) suits $R\sin(\theta + \alpha)$. $\cos\theta - \sqrt3\sin\theta$ suits $R\cos(\theta + \alpha)$, because $\cos(\theta + \alpha)$ has a minus. With the right choice, both $R\cos\alpha$ and $R\sin\alpha$ come out positive and $\alpha$ is acute.

2 Finding $R$ and $\alpha$

Expand the chosen form using the compound angle formula.
Compare coefficients of $\sin\theta$ and $\cos\theta$ to get $R\cos\alpha$ and $R\sin\alpha$.
Square and add: $R = \sqrt{a^2 + b^2}$.
Divide: $\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha}$.

The key results

$R = \sqrt{a^2 + b^2} \qquad\qquad \tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha}$

Worked Example 1 — $R\sin(x + \alpha)$ in degrees

Express $3\sin x + 4\cos x$ in the form $R\sin(x + \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$.

①$R\sin(x + \alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha$.

②Compare: $R\cos\alpha = 3$ and $R\sin\alpha = 4$.

③$R = \sqrt{3^2 + 4^2} = 5$. $\tan\alpha = \tfrac43 \Rightarrow \alpha = 53.13^\circ$ (2 d.p.).

④$3\sin x + 4\cos x \equiv 5\sin(x + 53.13^\circ)$.

$\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha}$, so here it is $\tfrac43$, not $\tfrac34$. Write down which coefficient matches $\sin\alpha$ before dividing.

Worked Example 2 — An exact form in radians

Express $\sqrt3\cos\theta + \sin\theta$ in the form $R\cos(\theta - \alpha)$, with $0 < \alpha < \tfrac{\pi}{2}$.

①$R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.

②$R\cos\alpha = \sqrt3$, $R\sin\alpha = 1$.

③$R = \sqrt{3 + 1} = 2$; $\tan\alpha = \tfrac{1}{\sqrt3} \Rightarrow \alpha = \tfrac{\pi}{6}$.

④$\sqrt3\cos\theta + \sin\theta \equiv 2\cos\left(\theta - \tfrac{\pi}{6}\right)$.

3 Maximum and Minimum Values

$\sin$ and $\cos$ always lie between $-1$ and $1$. So $R\sin(\theta + \alpha)$ has maximum $R$ and minimum $-R$, and you find where they happen by setting the bracket to the angle where $\sin$ (or $\cos$) is $\pm1$.

Worked Example 3 — Maximum, minimum and where they occur

Using Worked Example 1, find the maximum and minimum values of $3\sin x + 4\cos x$, and the values of $x$ between $0^\circ$ and $360^\circ$ where they occur.

①$5\sin(x + 53.13^\circ)$ has maximum $5$ when $x + 53.13^\circ = 90^\circ$, i.e. $x = 36.87^\circ$.

②Minimum $-5$ when $x + 53.13^\circ = 270^\circ$, i.e. $x = 216.87^\circ$.

Worked Example 4 — A reciprocal

Find the greatest and least values of $\dfrac{10}{15 + 3\cos x - 4\sin x}$.

①$3\cos x - 4\sin x = R\cos(x + \alpha)$ with $R\cos\alpha = 3$, $R\sin\alpha = 4$: $\;R = 5$.

②The denominator $15 + 5\cos(x + \alpha)$ ranges from $10$ to $20$.

③The fraction is greatest when the denominator is least: $\tfrac{10}{10} = 1$. Least value: $\tfrac{10}{20} = \tfrac12$.

A small denominator gives a large fraction. Swapping maximum and minimum here is a very common slip.

4 Solving Equations

To solve $a\cos\theta + b\sin\theta = c$: write the left side as a single function, then solve as in Year 1 — remembering to shift the interval by $\alpha$.

Worked Example 5 — Solving in degrees

Solve $3\sin x + 4\cos x = 2$ for $0^\circ \leqslant x \leqslant 360^\circ$.

①$5\sin(x + 53.13^\circ) = 2 \Rightarrow \sin(x + 53.13^\circ) = 0.4$.

②New interval: $53.13^\circ \leqslant x + 53.13^\circ \leqslant 413.13^\circ$.

③$\arcsin 0.4 = 23.58^\circ$ (too small), so use $180^\circ - 23.58^\circ = 156.42^\circ$ and $360^\circ + 23.58^\circ = 383.58^\circ$.

④Subtract $53.13^\circ$: $x = 103.3^\circ$ or $x = 330.4^\circ$ (1 d.p.).

Worked Example 6 — Exact solutions in radians

Solve $\cos\theta - \sqrt3\sin\theta = 1$ for $0 \leqslant \theta \leqslant 2\pi$.

①$R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha$: $R\cos\alpha = 1$, $R\sin\alpha = \sqrt3$.

②$R = 2$, $\tan\alpha = \sqrt3 \Rightarrow \alpha = \tfrac{\pi}{3}$. Equation: $2\cos\left(\theta + \tfrac{\pi}{3}\right) = 1$.

③$\cos\left(\theta + \tfrac{\pi}{3}\right) = \tfrac12$ with $\tfrac{\pi}{3} \leqslant \theta + \tfrac{\pi}{3} \leqslant \tfrac{7\pi}{3}$.

④$\theta + \tfrac{\pi}{3} = \tfrac{\pi}{3}, \tfrac{5\pi}{3}, \tfrac{7\pi}{3}$, so $\theta = 0, \tfrac{4\pi}{3}, 2\pi$.

Check $\theta = 0$: $\cos 0 - \sqrt3\sin 0 = 1$ ✓. Solutions at the ends of the interval are easy to miss.

Worked Example 7 — With a double angle

Show that $6\sin x\cos x + 8\cos^2 x$ can be written as $3\sin 2x + 4\cos 2x + 4$, and hence find its range.

①$6\sin x\cos x = 3\sin 2x$ and $8\cos^2 x = 4(1 + \cos 2x) = 4\cos 2x + 4$ ✓

②$3\sin 2x + 4\cos 2x = 5\sin(2x + 53.13^\circ)$, which ranges from $-5$ to $5$.

③So the expression ranges from $-1$ to $9$: $-1 \leqslant 6\sin x\cos x + 8\cos^2 x \leqslant 9$.

5 Quick Reference

Amplitude

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

Angle

$\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha}$

Choosing the form

Match signs so $\alpha$ is acute.

Max / min

$\pm R$, when the bracket gives $\sin$ or $\cos = \pm1$.

Reciprocals

Least denominator gives the greatest value.

Equations

Shift the interval by $\alpha$ before solving.

Double angles

$a\cos 2x + b\sin 2x$ works the same way.

Accuracy

Keep $\alpha$ to $2$ d.p. (degrees) or $3$–$4$ s.f. (radians).

6 Practice Questions

Question 1

Express $4\sin x - 3\cos x$ in the form $R\sin(x - \alpha)$, with $R > 0$ and $0^\circ < \alpha < 90^\circ$.

▶ Show solution

$R\sin(x - \alpha) = R\sin x\cos\alpha - R\cos x\sin\alpha$: $R\cos\alpha = 4$, $R\sin\alpha = 3$.

$R = 5$, $\tan\alpha = \tfrac34 \Rightarrow \alpha = 36.87^\circ$. $4\sin x - 3\cos x \equiv 5\sin(x - 36.87^\circ)$.

Question 2

Show that $\sin x + \cos x \equiv \sqrt2\sin\left(x + \tfrac{\pi}{4}\right)$.

▶ Show solution

$R\cos\alpha = 1$, $R\sin\alpha = 1$: $R = \sqrt2$, $\tan\alpha = 1 \Rightarrow \alpha = \tfrac{\pi}{4}$ ✓

Question 3

Find the maximum value of $7\cos\theta + 24\sin\theta$ and the smallest positive value of $\theta$ (in degrees) at which it occurs.

▶ Show solution

$R\cos(\theta - \alpha)$: $R\cos\alpha = 7$, $R\sin\alpha = 24$, $R = 25$, $\tan\alpha = \tfrac{24}{7} \Rightarrow \alpha = 73.74^\circ$.

Maximum $25$ when $\theta - \alpha = 0$, i.e. $\theta = 73.7^\circ$.

Question 4

Solve $\sin x + \cos x = 1$ for $0 \leqslant x \leqslant 2\pi$.

▶ Show solution

$\sqrt2\sin\left(x + \tfrac{\pi}{4}\right) = 1 \Rightarrow \sin\left(x + \tfrac{\pi}{4}\right) = \tfrac{\sqrt2}{2}$, with $\tfrac{\pi}{4} \leqslant x + \tfrac{\pi}{4} \leqslant \tfrac{9\pi}{4}$.

$x + \tfrac{\pi}{4} = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{9\pi}{4}$, so $x = 0, \tfrac{\pi}{2}, 2\pi$.

Question 5

Solve $2\cos\theta + 5\sin\theta = 3$ for $0^\circ \leqslant \theta \leqslant 360^\circ$, to $1$ decimal place.

▶ Show solution

$R\cos(\theta - \alpha)$: $R\cos\alpha = 2$, $R\sin\alpha = 5$, $R = \sqrt{29}$, $\alpha = \arctan\tfrac52 = 68.20^\circ$.

$\cos(\theta - 68.20^\circ) = \tfrac{3}{\sqrt{29}}$, with $-68.20^\circ \leqslant \theta - 68.20^\circ \leqslant 291.80^\circ$.

$\arccos\tfrac{3}{\sqrt{29}} = 56.15^\circ$, so $\theta - 68.20^\circ = 56.15^\circ$ or $-56.15^\circ$.

$\theta = 124.3^\circ$ or $12.1^\circ$.

Question 6

Express $5\cos x - 12\sin x$ in the form $R\cos(x + \alpha)$ with $0 < \alpha < \tfrac{\pi}{2}$, and hence state the greatest and least values of $\dfrac{1}{20 + 5\cos x - 12\sin x}$.

▶ Show solution

$R\cos\alpha = 5$, $R\sin\alpha = 12$: $R = 13$, $\alpha = \arctan\tfrac{12}{5} = 1.176$ rad (4 s.f.).

The denominator $20 + 13\cos(x + 1.176)$ ranges from $7$ to $33$.

Greatest value $\tfrac17$, least value $\tfrac{1}{33}$.

Question 7

Write $\sqrt3\sin x - \cos x$ in the form $R\sin(x - \alpha)$, and hence solve $\sqrt3\sin x - \cos x = \sqrt2$ for $0 \leqslant x \leqslant 2\pi$.

▶ Show solution

$R\cos\alpha = \sqrt3$, $R\sin\alpha = 1$: $R = 2$, $\alpha = \tfrac{\pi}{6}$.

$\sin\left(x - \tfrac{\pi}{6}\right) = \tfrac{\sqrt2}{2}$, with $-\tfrac{\pi}{6} \leqslant x - \tfrac{\pi}{6} \leqslant \tfrac{11\pi}{6}$.

$x - \tfrac{\pi}{6} = \tfrac{\pi}{4}$ or $\tfrac{3\pi}{4}$, so $x = \tfrac{5\pi}{12}$ or $\tfrac{11\pi}{12}$.

Question 8

Solve $3\cos 2x + 4\sin 2x = 2.5$ for $0 \leqslant x \leqslant \pi$, to $3$ significant figures.

▶ Show solution

$5\cos(2x - \alpha)$ with $\alpha = \arctan\tfrac43 = 0.9273$.

$\cos(2x - 0.9273) = 0.5$, with $-0.9273 \leqslant 2x - 0.9273 \leqslant 2\pi - 0.9273 = 5.356$.

$2x - 0.9273 = \tfrac{\pi}{3} = 1.0472$ or $2\pi - \tfrac{\pi}{3} = 5.2360$ ($-1.0472$ is outside the interval).

$2x = 1.9745$ or $6.1633$, so $x = 0.987$ or $3.08$.

Question 9

Find the range of $\mathrm{f}(x) = 2\sin x + 2\sqrt3\cos x - 1$.

▶ Show solution

$2\sin x + 2\sqrt3\cos x$: $R = \sqrt{4 + 12} = 4$.

So $\mathrm{f}(x) = 4\sin(x + \alpha) - 1$ (with $\alpha = \tfrac{\pi}{3}$), which ranges from $-5$ to $3$: $-5 \leqslant \mathrm{f}(x) \leqslant 3$.

Question 10

The temperature $T$ °C of a sample in a laboratory is modelled by $T = 20 + 3\cos\left(\dfrac{\pi t}{12}\right) + 4\sin\left(\dfrac{\pi t}{12}\right)$, where $t$ is the time in hours after midnight, $0 \leqslant t \leqslant 24$.

(a) Write $3\cos\left(\frac{\pi t}{12}\right) + 4\sin\left(\frac{\pi t}{12}\right)$ in the form $R\cos\left(\frac{\pi t}{12} - \alpha\right)$. (b) Find the maximum and minimum temperatures and the times at which they occur. (c) Find the times at which the temperature is $22$ °C.

▶ Show solution

(a) $R\cos\alpha = 3$, $R\sin\alpha = 4$: $R = 5$, $\alpha = \arctan\tfrac43 = 0.9273$ rad. $T = 20 + 5\cos\left(\tfrac{\pi t}{12} - 0.9273\right)$.

(b) Maximum $25$ °C when $\tfrac{\pi t}{12} = 0.9273$, i.e. $t = \dfrac{12 \times 0.9273}{\pi} = 3.54$ h (about 03:33).

Minimum $15$ °C when $\tfrac{\pi t}{12} - 0.9273 = \pi$, i.e. $t = 15.54$ h (about 15:33).

(c) $5\cos\left(\tfrac{\pi t}{12} - 0.9273\right) = 2 \Rightarrow \cos(\ldots) = 0.4$, with $-0.9273 \leqslant \tfrac{\pi t}{12} - 0.9273 \leqslant 5.356$.

$\arccos 0.4 = 1.1593$. Solutions in the interval: $1.1593$ and $2\pi - 1.1593 = 5.1239$ ($-1.1593$ is outside).

$t = \dfrac{12(1.1593 + 0.9273)}{\pi} = 7.97$ h and $t = \dfrac{12(5.1239 + 0.9273)}{\pi} = 23.11$ h — about 07:58 and 23:07.

The R cos(x + a) Form · A Level Maths Year 2 · Created with MathJax