Trigonometric Identities

Trigonometric identities

Consider the right-angled triangle shown in Figure 7.2.

Using Pythagoras' theorem:

$AB^2 + BC^2 = AC^2$

Since $AC = r$, $AB = x$, and $BC = y$, then:

$x^2 + y^2 = r^2$

Divide throughout by $r^2$:

$\frac{x^2}{r^2} + \frac{y^2}{r^2} = \frac{r^2}{r^2}$

$\left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = 1$

But $\cos \theta = \frac{x}{r}$ and $\sin \theta = \frac{y}{r}$.

Substituting the values for $\cos \theta$ and $\sin \theta$ gives:

$\cos^2 \theta + \sin^2 \theta = 1 \quad (7.1)$

Two similar identities can be deduced from equation (7.1) as follows:

Divide by $\sin^2 \theta$ both sides of equation (7.1):

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

whereas, $\cot \theta = \frac{\cos \theta}{\sin \theta}$

$\cot^2 \theta + 1 = \csc^2 \theta \quad (7.2)$

Again, divide by $\cos^2 \theta$ both sides of equation (7.1):

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

whereas, $\tan \theta = \frac{\sin \theta}{\cos \theta}$

$1 + \tan^2 \theta = \sec^2 \theta \quad (7.3)$

Equations (7.1), (7.2), and (7.3) are also known as Pythagorean identities as they express the Pythagorean theorem in terms of trigonometric functions.

Example 1

Simplify $\sin \theta + \cos \theta \cot \theta$.

Solution:

$\sin \theta + \cos \theta \cot \theta = \sin \theta + \cos \theta \frac{\cos \theta}{\sin \theta}$

$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta}$

Since $\sin^2 \theta + \cos^2 \theta = 1$,

$= \frac{1}{\sin \theta}$

$= \csc \theta$

Therefore, $\sin \theta + \cos \theta \cot \theta = \csc \theta$.

Example 2

If $x = \sin \theta$, then show that $\tan \theta = \frac{x}{\sqrt{1 - x^2}}$.

Solution:

Given $x = \sin \theta$

$\sin \theta = x$

$\sin^2 \theta = x^2$

Since $1 - \sin^2 \theta = \cos^2 \theta$,

$\cos^2 \theta = 1 - x^2$

$\cos \theta = \sqrt{1 - x^2}$

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{x}{\sqrt{1 - x^2}}$

Therefore, if $x = \sin \theta$, then $\tan \theta = \frac{x}{\sqrt{1 - x^2}}$.

Example 3

Solve the equation $2\sin^2 \theta - 1 = \cos \theta$, for values of $\theta$ between 0° and 360°.

Solution:

Given $2\sin^2 \theta - 1 = \cos \theta$

$2\sin^2 \theta - \cos \theta - 1 = 0$

But $\sin^2 \theta = 1 - \cos^2 \theta$, thus,

$2(1 - \cos^2 \theta) - \cos \theta - 1 = 0$

$2 - 2\cos^2 \theta - \cos \theta - 1 = 0$

$-2\cos^2 \theta - \cos \theta + 1 = 0$

$2\cos^2 \theta + \cos \theta - 1 = 0$

Factorizing the resulted quadratic equation gives:

$(2\cos \theta - 1)(\cos \theta + 1) = 0$

Hence $2\cos \theta - 1 = 0$ or $\cos \theta + 1 = 0$

$\cos \theta = \frac{1}{2} \quad \text{or} \quad \cos \theta = -1$

If $\cos \theta = \frac{1}{2}$, $\theta = 60^\circ, 300^\circ$.

If $\cos \theta = -1$, $\theta = 180^\circ$.

Therefore, the values of $\theta$ between 0° and 360° are 60°, 180°, and 300°.

Example 4

Solve the equation $4\cos \theta - 3\sec \theta = 2\tan \theta$, for $0^\circ \le \theta \le 180^\circ$.

Solution:

Given $4\cos \theta - 3\sec \theta = 2\tan \theta$

$4\cos \theta - \frac{3}{\cos \theta} = \frac{2\sin \theta}{\cos \theta}$

Multiplying both sides by $\cos \theta$ gives:

$4\cos^2 \theta - 3 = 2\sin \theta$

$4\cos^2 \theta - 2\sin \theta - 3 = 0$

But $\cos^2 \theta = 1 - \sin^2 \theta$

$4(1 - \sin^2 \theta) - 2\sin \theta - 3 = 0$

$4 - 4\sin^2 \theta - 2\sin \theta - 3 = 0$

$-4\sin^2 \theta - 2\sin \theta + 1 = 0$

$4\sin^2 \theta + 2\sin \theta - 1 = 0$

Using the quadratic formula for $\sin \theta$: