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
From Trigonometrical Ratios of 0° to HOME PAGE
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.
