# Trigonometrical Ratios of 0°

How to find the Trigonometrical Ratios of 0°?

Let a rotating line $$\overrightarrow{OX}$$ rotates about O in the anti clockwise sense and starting from its initial position $$\overrightarrow{OX}$$ traces out ∠XOY = θ where θ is very small.

Take a point P on $$\overrightarrow{OY}$$  and draw $$\overline{PQ}$$ perpendicular to $$\overrightarrow{OX}$$  .

Now according to the definition of trigonometric ratio we get,
sin θ = $$\frac{\overline{PQ}}{\overline{OP}}$$;
cos θ = $$\frac{\overline{OQ}}{\overline{OP}}$$ and
tan θ = $$\frac{\overline{PQ}}{\overline{OQ}}$$

When θ is slowly decreases and finally tends to zero then,
(a) $$\overline{PQ}$$ slowly decreases and finally tends to zero and

(b) the numerical difference between $$\overline{OP}$$  and $$\overline{OQ}$$  becomes very small and finally tends to zero.

Hence, in the Limit when θ → 00 then $$\overline{PQ}$$ → 0 and $$\overline{OP}$$   → $$\overline{OQ}$$  . Therefore, we get
$$\lim_{θ \to 0} sin θ = \lim_{θ \rightarrow 0}\frac{\overline{PQ}}{\overline{OP}} = \frac{0}{\overline{OQ}}$$ [since, θ → 0° therefore, $$\overline{PQ}$$ → 0].
= 0

Therefore sin 0° = 0

$$\lim_{θ \rightarrow 0} cos θ = \lim_{θ \rightarrow 0}\frac{\overline{OQ}}{\overline{OP}} = \frac{\overline{OQ}}{\overline{OQ}}$$, [since, θ → 0° therefore, $$\overline{OP}$$ → $$\overline{OQ}$$].
= 1

Therefore cos 0° = 1

$$\lim_{θ \rightarrow 0} tan θ = \lim_{θ \rightarrow 0}\frac{\overline{PQ}}{\overline{OQ}} = \frac{0}{\overline{OQ}}$$ [since, θ → 0° therefore, $$\overline{PQ}$$ → 0].
= 0

Therefore tan 0° = 0

Thus,
csc 0° = $$\frac{1}{sin 0°} = \frac{1}{0}$$, [since, sin 0° = 0]
= undefined

Therefore csc 0° = undefined

sec 0° = $$\frac{1}{cos 0°} = \frac{1}{1}$$, [since, cos 0° = 1]
= 1

Therefore sec 0° = 1

cot 0° = $$\frac{1}{tan 0°} = \frac{1}{0}$$, [since, tan 0° = 0]
= undefined

Therefore cot 0° = undefined

Trigonometrical Ratios of 0 degree are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.

Trigonometric Functions

### New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.