Trigonometric Formulae, Equations and Inequalities

Lessons

Senior 5 · Mathematics

Trigonometric Formulae, Equations and Inequalities

5 lessons

Lesson 4 of 5

The equation a sin x + b cos x = c and trigonometric inequalities

25 min

Two waves that make one

Draw $y = \sqrt{3}\sin x + \cos x$ and you see a single, smooth sine wave, only taller and shifted to the left (Figure 1). That is no accident. For any $R \gt 0$ and angle $\alpha$ , the addition formula gives

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

So $a\sin x + b\cos x$ equals $R\sin(x + \alpha)$ as soon as $R\cos\alpha = a$ and $R\sin\alpha = b$ . Squaring and adding gives $R^{2} = a^{2} + b^{2}$ ; dividing gives $\tan\alpha = \frac{b}{a}$ , with the quadrant of $\alpha$ fixed by the signs of $\cos\alpha = \frac{a}{R}$ and $\sin\alpha = \frac{b}{R}$ .

Figure 1.

\sqrt{3}\sin x + \cos x = 2\sin\left(x + \frac{\pi}{6}\right)

is a single sine wave of amplitude 2. It meets

y = 1

x = 0

and

x = \frac{2\pi}{3}

, and lies above the line (green) for

0 \lt x \lt \frac{2\pi}{3}

Definition

Harmonic form (amplitude–phase form) of $a\sin x + b\cos x$ , where $a$ and $b$ are not both 0:

a\sin x + b\cos x = R\sin(x + \alpha), \quad R = \sqrt{a^2 + b^2}, \quad \cos\alpha = \frac{a}{R}, \quad \sin\alpha = \frac{b}{R}.

$R$ is the amplitude and $\alpha$ the phase angle. The same expression can be written $R\cos(x - \beta)$ with $\cos\beta = \frac{b}{R}$ and $\sin\beta = \frac{a}{R}$ .

Example: $\sqrt{3}\sin x + \cos x$ : $R = \sqrt{3 + 1} = 2$ , $\cos\alpha = \frac{\sqrt{3}}{2}$ , $\sin\alpha = \frac{1}{2}$ , so $\alpha = \frac{\pi}{6}$ and the expression is $2\sin\left(x + \frac{\pi}{6}\right)$ .

Important

Solving $a\sin x + b\cos x = c$

Write the left side as $R\sin(x + \alpha)$ ; the equation becomes $\sin(x + \alpha) = \frac{c}{R}$ , a basic equation from Lesson 3.
Because $-1 \le \sin(x + \alpha) \le 1$ , there are solutions if and only if $|c| \le \sqrt{a^2 + b^2}$ .
The greatest value of $a\sin x + b\cos x$ is $R$ and the least is $- R$ .

Worked example 1: exact angles

Solve $\sqrt{3}\sin x + \cos x = 1$ for $0 \le x \lt 2\pi$ .

Solution From the definition box, the equation is $2\sin\left(x + \frac{\pi}{6}\right) = 1$ , so $\sin\left(x + \frac{\pi}{6}\right) = \frac{1}{2}$ :

x + \frac{\pi}{6} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{6} = \frac{5\pi}{6} + 2k\pi \quad \Rightarrow \quad x = 2k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + 2k\pi.

Answer: $x = 0$ or $x = \frac{2\pi}{3}$ , the two crossing points in Figure 1. Check: $\sqrt{3}\cdot\frac{\sqrt{3}}{2} + \left(-\frac{1}{2}\right) = \frac{3}{2} - \frac{1}{2} = 1$ . ✓

Worked example 2: a calculator angle

Solve $5\sin x - 12\cos x = 6.5$ for $0^\circ \le x \lt 360^\circ$ , and find the greatest value of the left side.

Solution Here a minus sign suits the form $R\sin(x - \alpha) = R\cos\alpha\sin x - R\sin\alpha\cos x$ . So $R\cos\alpha = 5$ , $R\sin\alpha = 12$ , $R = \sqrt{25 + 144} = 13$ and $\alpha = \tan^{-1}\frac{12}{5} = 67.38^\circ$ (first quadrant, as both are positive).

13\sin(x - 67.38^\circ) = 6.5 \ \Rightarrow \ \sin(x - 67.38^\circ) = 0.5 \ \Rightarrow \ x - 67.38^\circ = 30^\circ \ \text{or} \ 150^\circ.

Answer: $x = 97.38^\circ$ or $x = 217.38^\circ$ . Solutions exist because $6.5 \le 13$ . The greatest value is $13$ , when $x - 67.38^\circ = 90^\circ$ , that is $x = 157.38^\circ$ .

A second method: the substitution $t = \tan\frac{x}{2}$

With the t-formulae of Lesson 2, $a\sin x + b\cos x = c$ becomes $\frac{2at + b(1 - t^2)}{1 + t^2} = c$ , a quadratic equation in $t$ . This method needs no phase angle, but $t = \tan\frac{x}{2}$ does not exist when $x = \pi + 2k\pi$ . Those values are never produced by the quadratic, so you must test $x = \pi$ in the original equation yourself.

Worked example 3: by the t-substitution

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

Solution

\frac{2t}{1 + t^2} + \frac{2(1 - t^2)}{1 + t^2} = 1 \ \Rightarrow \ 2t + 2 - 2t^2 = 1 + t^2 \ \Rightarrow \ 3t^2 - 2t - 1 = 0 \ \Rightarrow \ (3t + 1)(t - 1) = 0.

$t = 1$ : $\frac{x}{2} = \frac{\pi}{4} + k\pi$ , so $x = \frac{\pi}{2}$ .
$t = -\frac{1}{3}$ : then $\sin x = \frac{2t}{1 + t^2} = -\frac{3}{5}$ and $\cos x = \frac{1 - t^2}{1 + t^2} = \frac{4}{5}$ , a fourth-quadrant angle: $x = 2\pi - \tan^{-1}\frac{3}{4} \approx 5.640$ (about $323.13^\circ$ ).
Test $x = \pi$ : $\sin\pi + 2\cos\pi = -2 \ne 1$ , so nothing was lost.

Check: $-\frac{3}{5} + 2\cdot\frac{4}{5} = 1$ . ✓

Warning

The t-substitution can hide $x = \pi$ . For $\sin x + \cos x = -1$ it gives $2 t + 1 - t^{2} = - 1 - t^{2}$ , so $t = - 1$ and $x = \frac{3\pi}{2}$ only. But $x = \pi$ is also a solution: $0 + (- 1) = - 1$ . Always test $x = \pi$ separately.

Trigonometric inequalities

To solve an inequality such as $\cos x \ge -\frac{\sqrt{2}}{2}$ , use the equation to find the boundary, and the graph (or the unit circle) to choose the side:

Solve the equation $\cos x = -\frac{\sqrt{2}}{2}$ in one period: $x = \frac{3\pi}{4}$ or $\frac{5\pi}{4}$ .
Decide on which arcs between the boundary points the curve is above or below the line.
Write the answer as a union of intervals, closed at the ends for $\leq \le$ or $\geq \ge$ , open for $< \lt$ or $> \gt$ . For the general solution, add $2k\pi$ (or $k\pi$ for a tangent) to every interval.

Figure 2.

\cos x \ge -\frac{\sqrt{2}}{2}

on the green arcs:

x \in \left[0, \frac{3\pi}{4}\right] \cup \left[\frac{5\pi}{4}, 2\pi\right)

. The boundary points come from the equation

\cos x = -\frac{\sqrt{2}}{2}

Definition

Trigonometric inequality: an inequality in which the unknown appears inside a trigonometric function, such as $\sin x \lt \frac{1}{2}$ or $\tan 2x \gt 1$ . Its solution is a set of intervals, usually infinitely many, repeating with the period of the function.

Example: $\sin x \gt 0$ for $x \in (0, \pi)$ , and in general for $x \in (2k\pi, \pi + 2k\pi)$ , $k \in \mathbb{Z}$ .

Worked example 4: a tangent of a double angle

Solve $\tan 2x \gt 1$ for $0 \lt x \lt \pi$ .

Solution On one branch of the tangent, $-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$ , the tangent increases from $-\infty$ to $+\infty$ , so $\tan\theta \gt 1$ exactly when $\frac{\pi}{4} \lt \theta \lt \frac{\pi}{2}$ . With the period $π \pi$ : $\frac{\pi}{4} + k\pi \lt 2x \lt \frac{\pi}{2} + k\pi$ , so

\frac{\pi}{8} + \frac{k\pi}{2} \lt x \lt \frac{\pi}{4} + \frac{k\pi}{2}.

$k = 0$ and $k = 1$ give the answer in $(0, \pi)$ : $x \in \left(\frac{\pi}{8}, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{8}, \frac{3\pi}{4}\right)$ . The right ends are open because $\tan 2x$ is undefined there.

Worked example 5: a table of signs

Solve $\sin 2x \gt \cos x$ for $0 \le x \lt 2\pi$ .

Solution $\sin 2x - \cos x = 2\sin x\cos x - \cos x = \cos x\,(2\sin x - 1)$ , and we need this product to be positive. The factors are zero at $x = \frac{\pi}{2}, \frac{3\pi}{2}$ (for $\cos x$ ) and $x = \frac{\pi}{6}, \frac{5\pi}{6}$ (for $2\sin x - 1$ ):

$x$	$\left[0, \frac{\pi}{6}\right)$	$\left(\frac{\pi}{6}, \frac{\pi}{2}\right)$	$\left(\frac{\pi}{2}, \frac{5\pi}{6}\right)$	$\left(\frac{5\pi}{6}, \frac{3\pi}{2}\right)$	$\left(\frac{3\pi}{2}, 2\pi\right)$
$\cos x$	$+$	$+$	$-$	$-$	$+$
$2\sin x - 1$	$-$	$+$	$+$	$-$	$-$
product	$-$	$+$	$-$	$+$	$-$

Answer: $x \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right) \cup \left(\frac{5\pi}{6}, \frac{3\pi}{2}\right)$ . Test one point: at $x = \pi$ , $\sin 2\pi = 0 \gt \cos\pi = -1$ . ✓

Tip

Inequalities in harmonic form. Write $a\sin x + b\cos x$ as $R\sin(x + \alpha)$ first, then solve for the angle $x + \alpha$ . For example, Figure 1 shows at once that $\sqrt{3}\sin x + \cos x \gt 1$ for $0 \lt x \lt \frac{2\pi}{3}$ in $[0, 2\pi)$ .

A Level Maths Tutor (John Armstrong): The Form Rsin(x+a) and Rcos(x+a) Explained From Basics. Watch how R and α come from matching coefficients, and how the form gives the greatest and least values and the solutions of a sin x + b cos x = c.

✦

Practice

Angles in degrees to two decimal places where they are not exact. The last two are harder.

Write $4\sin x + 3\cos x$ in the form $R\sin(x + \alpha)$ with $0^\circ \lt \alpha \lt 90^\circ$ . State its greatest value and the $x$ in $[0^\circ, 360^\circ)$ where it occurs.
Solve $\sin x - \cos x = 1$ for $0 \le x \lt 2\pi$ .
Show that $2\sin x + 5\cos x = 6$ has no real solution.
Use $t = \tan\frac{x}{2}$ to solve $2\sin x + \cos x = 2$ for $0 \le x \lt 2\pi$ .
Solve $\sin x \lt -\frac{1}{2}$ for $0 \le x \lt 2\pi$ .
Solve $\cos 2x \ge \frac{1}{2}$ for $0 \le x \lt \pi$ .
Solve $\tan x \le \sqrt{3}$ for $-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$ .
Solve $\sin x - \sqrt{3}\cos x \ge 1$ for $0 \le x \lt 2\pi$ .

Show answersHide answers

$R = 5$ , $\tan\alpha = \frac{3}{4}$ , $\alpha = 36.87^\circ$ : $5\sin(x + 36.87^\circ)$ . Greatest value 5, when $x + 36.87^\circ = 90^\circ$ , so $x = 53.13^\circ$ .
$\sqrt{2}\sin\left(x - \frac{\pi}{4}\right) = 1$ , so $x - \frac{\pi}{4} = \frac{\pi}{4}$ or $\frac{3\pi}{4}$ : $x = \frac{\pi}{2}$ or $π \pi$ .
$R = \sqrt{4 + 25} = \sqrt{29} \approx 5.39 \lt 6$ , so the left side never reaches 6.
$4 t + 1 - t^{2} = 2 + 2 t^{2}$ , so $3 t^{2} - 4 t + 1 = 0$ : $t = 1$ gives $x = \frac{\pi}{2}$ ; $t = \frac{1}{3}$ gives $x = 2\tan^{-1}\frac{1}{3} \approx 0.644$ ( $36.87^\circ$ ). $x = \pi$ gives $-1 \ne 2$ .
$\sin x = -\frac{1}{2}$ at $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ ; below the line between them: $x \in \left(\frac{7\pi}{6}, \frac{11\pi}{6}\right)$ .
$2x \in \left[0, \frac{\pi}{3}\right] \cup \left[\frac{5\pi}{3}, 2\pi\right)$ , so $x \in \left[0, \frac{\pi}{6}\right] \cup \left[\frac{5\pi}{6}, \pi\right)$ .
The tangent increases on this branch and $\tan\frac{\pi}{3} = \sqrt{3}$ : $x \in \left(-\frac{\pi}{2}, \frac{\pi}{3}\right]$ .
$2\sin\left(x - \frac{\pi}{3}\right) \ge 1$ , so $\frac{\pi}{6} \le x - \frac{\pi}{3} \le \frac{5\pi}{6}$ : $x \in \left[\frac{\pi}{2}, \frac{7\pi}{6}\right]$ .

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the table of signs, the practice items and the quiz questions.
Video: Trigonometry: The Form Rsin(x+a) and Rcos(x+a) Explained From Basics, A Level Maths Tutor (John Armstrong), embedded from YouTube.

1Two waves that make one

2A second method: the substitution t=tan⁡x2 t = \tan\frac{x}{2} t=tan2x​

3Trigonometric inequalities

Two waves that make one

A second method: the substitution $t = \tan\frac{x}{2}$

Trigonometric inequalities