Trigonometric equations of the form $a\cos\theta + b\sin\theta = c$

Trigonometric equations of the form $a\cos\theta + b\sin\theta = c$, where a, b, and c are real numbers, can be solved by using t-formulae or by expressing it in the form of $R\cos(\theta \pm \alpha) = c$ or $R\sin(\theta \pm \alpha) = c$.

Using t-formulae

Recall the double angle formulae:

1. $\sin 2x = 2\sin x \cos x$
2. $\cos 2x = \cos^2 x - \sin^2 x$

Both equations for $\sin 2x$ and $\cos 2x$ can be expressed in terms of $\tan \frac{\theta}{2}$.

From $\sin 2x = 2\sin x \cos x$:

$\sin 2x = \frac{2\sin x \cos x}{1} = \frac{2\sin x \cos x}{\cos^2 x + \sin^2 x} = \frac{\frac{2\sin x}{\cos x}}{1 + \frac{\sin^2 x}{\cos^2 x}} = \frac{2\tan x}{1 + \tan^2 x}$

Therefore, $\sin 2x = \frac{2\tan x}{1 + \tan^2 x}$ (7.7)

Now, let $\theta = 2x$, then $x = \frac{\theta}{2}$, equation (7.7) becomes:

$\sin \theta = \frac{2\tan\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}}$

Let $t = \tan\frac{\theta}{2}$, then $\sin \theta = \frac{2t}{1 + t^2}$. This is called the t-formula for $\sin\theta$.

Example 1

Solve the equation $5\cos \theta - 2\sin \theta = 2$ for $0^\circ \le \theta \le 360^\circ$ by using the substitution $t = \tan\frac{\theta}{2}$.

Solution:

Substituting the t-formulae into the equation:

$5\left(\frac{1 - t^2}{1 + t^2}\right) - 2\left(\frac{2t}{1 + t^2}\right) = 2$

$5(1 - t^2) - 4t = 2(1 + t^2)$

$5 - 5t^2 - 4t = 2 + 2t^2$

$7t^2 + 4t - 3 = 0$

$(7t - 3)(t + 1) = 0$

$t = \frac{3}{7} \text{ or } t = -1$

If $t = \frac{3}{7}$, $\tan\frac{\theta}{2} = \frac{3}{7}$, so $\frac{\theta}{2} \approx 23.2^\circ$, and $\theta \approx 46.4^\circ$ or $180^\circ + 46.4^\circ = 226.4^\circ$

If $t = -1$, $\tan\frac{\theta}{2} = -1$, so $\frac{\theta}{2} = -45^\circ$ or $135^\circ$, and $\theta = -90^\circ$ or $270^\circ$.

Therefore, the solutions for $0^\circ \le \theta \le 360^\circ$ are approximately $46.4^\circ$ and $270^\circ$.

Expressing $a\cos\theta + b\sin\theta$ in the form $R\cos(\theta \pm \alpha)$ or $R\sin(\theta \pm \alpha)$

The technique of expressing $a\cos\theta + b\sin\theta$ in the form of $R\cos(\theta \pm \alpha)$ or $R\sin(\theta \pm \alpha)$ is useful in finding maximum and minimum values of trigonometric equations of the form $a\cos\theta + b\sin\theta = c$.

Expressing $a\cos\theta + b\sin\theta$ in the form of $R\cos(\theta - \alpha)$

Using the compound angle formula for $R\cos(\theta - \alpha)$ gives:

$R\cos(\theta - \alpha) \equiv a\cos\theta + b\sin\theta \equiv R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$

Comparing the coefficients of $\cos\theta$ gives:

$a = R\cos\alpha$ ………………………………………… (i)

Comparing the coefficients of $\sin\theta$ gives:

$b = R\sin\alpha$ ………………………………………… (ii)

Squaring and adding equations (i) and (ii) gives:

$a^2 + b^2 = R^2\cos^2\alpha + R^2\sin^2\alpha$

$a^2 + b^2 = R^2(\cos^2\alpha + \sin^2\alpha)$

$R^2 = a^2 + b^2$

$R = \sqrt{a^2 + b^2}$ (where R is positive)

The value of $\alpha$ is obtained by dividing equation (ii) by equation (i):

$\tan\alpha = \frac{R\sin\alpha}{R\cos\alpha} = \frac{b}{a}$

$\alpha = \tan^{-1}\left(\frac{b}{a}\right)$

Example 2

Express $3\cos\theta - 4\sin\theta$ in the form of $R\cos(\theta - \alpha)$.

Solution:

Let $3\cos\theta - 4\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$

Comparing coefficients:

$R\cos\alpha = 3$

$R\sin\alpha = -4$

$R = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$

$\tan\alpha = \frac{-4}{3}$

$\alpha = \tan^{-1}\left(\frac{-4}{3}\right) \approx -53.1^\circ$ (or 306.9°)

Therefore, $3\cos\theta - 4\sin\theta = 5\cos(\theta + 53.1^\circ)$.

Example 3

Show that $5\cos\theta - 12\sin\theta$ can be expressed in the form $13\cos(\theta + 67.4^\circ)$.

Solution:

Let $5\cos\theta - 12\sin\theta = R\cos(\theta + \alpha)$

Comparing coefficients:

$R\cos\alpha = 5$

$R\sin\alpha = -12$

$R = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13$

$\tan\alpha = \frac{-12}{5}$

$\alpha \approx -67.4^\circ$ (or 292.6°)

Therefore, $5\cos\theta - 12\sin\theta = 13\cos(\theta + 67.4^\circ)$.

Expressing $a\cos\theta + b\sin\theta$ in the form $R\sin(\theta + \alpha)$

Using the compound angle formula for $R\sin(\theta + \alpha)$ gives:

$a\cos\theta + b\sin\theta = R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$

Comparing coefficients:

$b = R\cos\alpha$ ………………………………………… (i)

$a = R\sin\alpha$ ………………………………………… (ii)

Squaring and adding equations (i) and (ii) gives:

$a^2 + b^2 = R^2$

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

Dividing equation (ii) by equation (i) gives:

$\tan\alpha = \frac{a}{b}$

$\alpha = \tan^{-1}\left(\frac{a}{b}\right)$

Example 4

Express $7\cos\theta + 24\sin\theta$ in the form of $R\sin(\theta + \alpha)$.

Solution:

Let $7\cos\theta + 24\sin\theta = R\sin(\theta + \alpha)$

Comparing coefficients:

$R\cos\alpha = 24$

$R\sin\alpha = 7$

$R = \sqrt{7^2 + 24^2} = 25$

$\tan\alpha = \frac{7}{24}$

$\alpha \approx 16.3^\circ$

Therefore, $7\cos\theta + 24\sin\theta = 25\sin(\theta + 16.3^\circ)$.

Example 5

Find the maximum and minimum values of $24\sin\theta - 7\cos\theta$, stating the values of $\theta$ for which the maximum and minimum values are attained.

Solution:

Let $24\sin\theta - 7\cos\theta = R\sin(\theta - \alpha)$

$R = \sqrt{24^2 + (-7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25$

$\tan\alpha = \frac{-7}{24}$

$\alpha \approx -16.3^\circ$

So, $24\sin\theta - 7\cos\theta = 25\sin(\theta + 16.3^\circ)$

Since $-1 \le \sin(\theta + 16.3^\circ) \le 1$, the maximum value is 25 (when $\sin(\theta + 16.3^\circ) = 1$) and the minimum value is -25 (when $\sin(\theta + 16.3^\circ) = -1$).

For the maximum value:

$\sin(\theta + 16.3^\circ) = 1$

$\theta + 16.3^\circ = 90^\circ$

$\theta = 73.7^\circ$

For the minimum value:

$\sin(\theta + 16.3^\circ) = -1$

$\theta + 16.3^\circ = 270^\circ$

$\theta = 253.7^\circ$

Therefore, the maximum value is 25 at $\theta = 73.7^\circ$, and the minimum value is -25 at $\theta = 253.7^\circ$.

Example 6

Show that $13\cos\lambda + 7\sin\lambda$ can be expressed in the form $\sqrt{218}\cos(\lambda - \alpha)$, where $\tan\alpha = \frac{7}{13}$. Hence, find the maximum and minimum values of the function, giving the corresponding values of $\lambda$.

Solution:

Let $13\cos\lambda + 7\sin\lambda = R\cos(\lambda - \alpha)$

$R = \sqrt{13^2 + 7^2} = \sqrt{169 + 49} = \sqrt{218}$

$\tan\alpha = \frac{7}{13}$

$\alpha \approx 28.3^\circ$

So, $13\cos\lambda + 7\sin\lambda = \sqrt{218}\cos(\lambda - 28.3^\circ)$

The maximum value is $\sqrt{218}$ (when $\cos(\lambda - 28.3^\circ) = 1$), and the minimum value is $-\sqrt{218}$ (when $\cos(\lambda - 28.3^\circ) = -1$).

For the maximum value:

$\lambda - 28.3^\circ = 0^\circ$

$\lambda = 28.3^\circ$

For the minimum value:

$\lambda - 28.3^\circ = 180^\circ$

$\lambda = 208.3^\circ$

Example 7

Solve the equation $5\sqrt{2} = 4\cos 2x + 3\sin 2x$ for values in the interval $-180^\circ \le x \le 180^\circ$.

Solution:

Let $4\cos 2x + 3\sin 2x = R\sin(2x + \alpha)$

$R = \sqrt{4^2 + 3^2} = 5$

$\tan\alpha = \frac{4}{3}$

$\alpha \approx 53.1^\circ$

So, $5\sqrt{2} = 5\sin(2x + 53.1^\circ)$

$\sin(2x + 53.1^\circ) = \sqrt{2}$

Since the maximum value of sine is 1, and $\sqrt{2} > 1$, there are no solutions to this equation.

Example 8

Solve the equation $3\cos x + \sin x = 2$ for $0^\circ \le x \le 360^\circ$.

Solution:

Let $3\cos x + \sin x = R\cos(x - \alpha)$

$R = \sqrt{3^2 + 1^2} = \sqrt{10}$

$\tan\alpha = \frac{1}{3}$

$\alpha \approx 18.4^\circ$

So, $\sqrt{10}\cos(x - 18.4^\circ) = 2$

$\cos(x - 18.4^\circ) = \frac{2}{\sqrt{10}}$

$x - 18.4^\circ \approx \pm 50.8^\circ$

$x \approx 69.2^\circ \text{ or } x \approx -32.4^\circ + 360^\circ = 327.6^\circ$

Therefore, the solutions are approximately $x = 69.2^\circ$ and $327.6^\circ$.

General solutions of trigonometric equations

A general solution is a solution that contains all possible solutions of a given trigonometric equation.

Consider the graph of the straight lines $y = a$, where $-1 \le a \le 1$, and $y = \cos \theta$ as shown in Figure 7.5:

The principal value of the equation $\cos \theta = a$ is $\alpha = \cos^{-1} a$. Since the cosine function is periodic with a period of $360^\circ$ or $2\pi$ radians, other solutions are obtained by adding or subtracting multiples of $360^\circ$ or $2\pi$ to $\pm\alpha$.

The general solution of the equation $\cos\theta = a$ where $-1 \le a \le 1$ is given as:

In degrees: $\theta = \pm\alpha + 360^\circ n$, where $\alpha = \cos^{-1} a$ and n is an integer.

In radians: $\theta = \pm\alpha + 2\pi n$, where $\alpha = \cos^{-1} a$ and n is an integer.

Similarly, the general solutions for sine and tangent are:

In degrees:

1. If $\sin \theta = \sin \alpha$, then $\theta = n180^\circ + (-1)^n \alpha$
2. If $\tan \theta = \tan \alpha$, then $\theta = \alpha + 180^\circ n$

In radians:

1. If $\sin \theta = \sin \alpha$, then $\theta = n\pi + (-1)^n \alpha$
2. If $\tan \theta = \tan \alpha$, then $\theta = \alpha + n\pi$

where n is an integer.

Example 1

Find the general solution of the equation $\cos x + \sin x = 1$, giving your answer in terms of $\pi$.

Solution:

Express $\cos x + \sin x$ in the form $R\cos(x - \alpha)$:

$R = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\tan\alpha = \frac{1}{1} = 1, \text{ so } \alpha = \frac{\pi}{4}$

The equation becomes: $\sqrt{2}\cos(x - \frac{\pi}{4}) = 1$

$\cos(x - \frac{\pi}{4}) = \frac{1}{\sqrt{2}}$

$x - \frac{\pi}{4} = \pm\frac{\pi}{4} + 2n\pi$

Case 1: $x - \frac{\pi}{4} = \frac{\pi}{4} + 2n\pi$

$x = \frac{\pi}{2} + 2n\pi$

Case 2: $x - \frac{\pi}{4} = -\frac{\pi}{4} + 2n\pi$

$x = 2n\pi$

Therefore, the general solutions are $x = \frac{\pi}{2} + 2n\pi$ and $x = 2n\pi$, where n is an integer.

Example 2

Find the general solution of the equation $\cot x + \csc x = \sqrt{3}$, giving the answer in terms of $\pi$.

Solution:

$\frac{\cos x}{\sin x} + \frac{1}{\sin x} = \sqrt{3}$

$\cos x + 1 = \sqrt{3}\sin x$

$\sqrt{3}\sin x - \cos x = 1$

Express $\sqrt{3}\sin x - \cos x$ in the form $R\sin(x - \alpha)$:

$R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$

$\tan\alpha = \frac{-1}{\sqrt{3}}, \text{ so } \alpha = -\frac{\pi}{6} \text{ or } \frac{5\pi}{6}.$ Using the positive value is easier for this example.

The equation becomes: $2\sin(x - \frac{5\pi}{6}) = 1$

$\sin(x - \frac{5\pi}{6}) = \frac{1}{2}$

$x - \frac{5\pi}{6} = n\pi + (-1)^n \frac{\pi}{6}$

$x = n\pi + (-1)^n \frac{\pi}{6} + \frac{5\pi}{6}$

Example 3

Use the t-formulae to find the general solution of the equation $2\cos x - \sin x = 1$, giving the answers in radians.

Solution:

Using the t-formulae, where $t = \tan\frac{x}{2}$:

$2\left(\frac{1 - t^2}{1 + t^2}\right) - \frac{2t}{1 + t^2} = 1$

$2 - 2t^2 - 2t = 1 + t^2$

$3t^2 + 2t - 1 = 0$

$(3t - 1)(t + 1) = 0$

$t = \frac{1}{3} \text{ or } t = -1$

Case 1: $\tan\frac{x}{2} = \frac{1}{3}$

$\frac{x}{2} = \arctan\left(\frac{1}{3}\right) + n\pi$

$x = 2\arctan\left(\frac{1}{3}\right) + 2n\pi \approx 0.6435 + 2n\pi$

Case 2: $\tan\frac{x}{2} = -1$

$\frac{x}{2} = -\frac{\pi}{4} + n\pi$

$x = -\frac{\pi}{2} + 2n\pi$

Therefore, the general solutions are $x \approx 0.6435 + 2n\pi$ and $x = -\frac{\pi}{2} + 2n\pi$, where n is an integer.