# Introduction to Trig Ratios

The major functions of trigonometric ratios are sine, cosine, tangent, cosecant, secant and cotangent.

These functions are used to find the relationship between the angle and side of a right angled triangle.

Some specific angles are:

• 0° and 90°
• 45°
• 30° and 60°

## Trigonometric Ratios of 0° and 90°

In ΔABC, right – angled at B, and ∠BAC = θ

So, from ΔABC, we have

Sin θ = BC / AC
Cos θ = AB / AC
Tan θ = BC / AB

### Case I: ∠A is becoming small

If ∠A is made smaller and smaller in the ΔABC, till it becomes zero. As ∠A gets smaller and smaller, the length of the BC decreases. The point C gets closer to point B, and finally when A becomes very close to 0°, AC becomes almost the same as AB.

When ∠A is very close to 0°, BC gets very close to 0 and so the value of

Sin A = BC / AC is very close to 0.

Also, when A is very close to 0°, AC is same as AB and so the value of

Cos A = AB/AC is very close to 1.

From the above discussion, we have

Sin 0° = 0
Cosec 0° = 1 / Sin 0° = 1/0 = not defined
Cosec 0° = ∞
Cos 0° = 1
Sec 0° = 1 / Cos 0° = 1/1 = 1
Sec 0° = 1

Using Sin and Cos values, we can find Tan 0°

Tan 0° = Sin 0° / Cos 0° = 0
Tan 0° = 0

Also, Cot 0° = 1 / Tan 0° = 1/0 = not defined

Cot 0° = ∞

### Case II: ∠A is becoming large

Now, let’s see when ∠A is made larger and larger in ΔABC till it becomes 90°. As ∠A gets larger and larger, ∠C gets smaller and smaller. So, the length of the side AB goes on decreasing. The point A gets closer to point B. Finally when ∠A is very close to 90°, ∠C becomes very close to 0° and the side AC almost coincides with side BC.

When ∠C is very close to 0°, A is very close to 90°, side AC is nearly the same as side BC.

So, Sin A is very close to 1.

From above discussion we get,

Sin 90° = 1
Cosec 90° = 1
Cos 90°= 0
Sec 90° = ∞
Tan 90°= ∞
Cot 90° = 0

## Trigonometric Ratios of 45°

In ΔABC, right angled at B, if one angle is 45, then the other angle by angle sum property of triangle will also be 45.
∠A = ∠C = 45°

So, BC = AB (Isosceles triangle property)

Let, AB = BC = ‘a’

Then by Pythagoras theorem, AC2 = AB2 + BC2

AC2 = a2 + a2 = 2a2

AC = a√2.

Using formulas for trigonometric ratios:

Sin 45° = Side opposite to angle 45°    = a/a√2 = 1/√2
Hypotenuse

Cos 45° = Side adjacent to angle 45°    = a/a√2 = 1/√2

Hypotenuse

Tan 45° = Side opposite to angle 45°     = a/a = 1

Also, Cosec 45° = √2, Sec 45° = √2, Cot 45° = 1

## Trigonometric Ratios of 30° and 60°

Let ΔABC, be an equilateral triangle. So, ∠A = ∠B = ∠C = 60°

Draw perpendicular AD from A to the side BC.

Now, ΔABD ≅ ΔACD (by ASA)

Therefore, BD = DC

Consider, ΔABD

A = 30, B = 60, D = 90

Let AB = x

So, BD = x/2

And we will find the length of AD by Pythagoras theorem.

Using formulas for trigonometric ratios:

Sin 30° = Side opposite to angle 30°    =x/2 / x = 1/2
Hypotenuse

Cos 30° = Side adjacent to angle 30°    = x√3/2 / x = √3/2

Hypotenuse

Tan 30° = Side opposite to angle 30°     = x/2 / x√3/2 = 1/√3

Also, Cosec 30° = 2, Sec 30° = 2/√3, Cot 30° = √3

Similarly,

Sin 60° = √3/2

Cosec 60° = 2/√3

Cos 60° = ½

Sec 60° 2

Tan 60° = √3

Cot 60° = 1/√3

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