Trigonometric ratios

Trigonometry

Trigonometry is a branch of mathematics that deals with relationships between angles and sides of triangles.

Trigonometric Ratios

The basic three trigonometrical ratios are sine, cosine, and tangent, which are written in short as sin, cos, and tan respectively.

Consider a right-angled triangle:

$$\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}, \quad \cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}}, \quad \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$$

Trigonometric Ratios in the Unit Circle

We can define trigonometric ratios for all angles using a unit circle centered at the origin (0,0).

If $90^\circ < \theta < 180^\circ$, the trigonometrical ratios are related to $180^\circ - \theta$.

If $180^\circ < \theta < 270^\circ$, the trigonometrical ratios are related to $\theta - 180^\circ$.

If $270^\circ < \theta < 360^\circ$, the trigonometrical ratios are related to $360^\circ - \theta$.

The signs (positive or negative) of the trigonometrical ratios depend on the quadrant in which the angle lies, summarized in the ASTC (All Students Take Calculus) diagram:

1st Quadrant (0°–90°): All ratios are positive.

2nd Quadrant (90°–180°): Sine is positive.

3rd Quadrant (180°–270°): Tangent is positive.

4th Quadrant (270°–360°): Cosine is positive.

Trigonometric Ratios to Solve Problems in Daily Life

Example 1

Write the signs of the following ratios:

$\sin(170^\circ)$

$\cos(240^\circ)$

$\tan(310^\circ)$

$\sin(300^\circ)$

Solution:

1. $\sin(170^\circ)$: 170° is in the 2nd quadrant where sine is positive. Thus, $\sin(170^\circ) = \sin(180^\circ - 170^\circ) = \sin(10^\circ)$ (positive)

2. $\cos(240^\circ)$: 240° is in the 3rd quadrant where cosine is negative. Thus, $\cos(240^\circ) = -\cos(240^\circ - 180^\circ) = -\cos(60^\circ)$

3. $\tan(310^\circ)$: 310° is in the 4th quadrant where tangent is negative. Thus, $\tan(310^\circ) = -\tan(360^\circ - 310^\circ) = -\tan(50^\circ)$

4. $\sin(300^\circ)$: 300° is in the 4th quadrant where sine is negative. Thus, $\sin(300^\circ) = -\sin(360^\circ - 300^\circ) = -\sin(60^\circ)$

Relationship between Trigonometric Ratios

The sine of an angle is equal to the cosine of its complement:

$$\sin(\theta) = \cos(90^\circ - \theta) \quad \text{and} \quad \cos(\theta) = \sin(90^\circ - \theta)$$

Also, from Pythagoras' Theorem in a right-angled triangle:

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

Example 2

Given that $\cos(A) = 0.8$ and $A$ is an acute angle, find:

1. $\sin(A)$
2. $\tan(A)$

Solution:

Using the identity:

$$\sin^2(A) + \cos^2(A) = 1$$

Thus,

$$\sin^2(A) = 1 - \cos^2(A) = 1 - (0.8)^2 = 1 - 0.64 = 0.36$$

Thus,

$$\sin(A) = \sqrt{0.36} = 0.6$$

Now,

$$\tan(A) = \frac{\sin(A)}{\cos(A)} = \frac{0.6}{0.8} = 0.75$$

Example 3

Given that $\theta$ and $\beta$ are acute angles such that:

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

and that:

$$\sin(\theta) = 0.6$$

Find $\tan(\beta)$.

Solution:

Since $\theta + \beta = 90^\circ$, $\beta = 90^\circ - \theta$.

Thus,

$$\sin(\theta) = \cos(\beta)$$

Thus:

$$\cos(\beta) = 0.6$$

Now, using the identity:

$$\sin^2(\beta) + \cos^2(\beta) = 1$$

Thus,

$$\sin^2(\beta) = 1 - (0.6)^2 = 1 - 0.36 = 0.64$$

Thus,

$$\sin(\beta) = \sqrt{0.64} = 0.8$$

Therefore,

$$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)} = \frac{0.8}{0.6} = \frac{4}{3}$$