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.

11 and 12 Grade Math

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