Trigonometrical Ratios of 45°

How to find the trigonometrical Ratios of 45°?

Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anti-clockwise sense and starting from the initial position \(\overrightarrow{OX}\) traces out ∠AOB = 45°.

Trigonometrical Ratios of 45°

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

Now, ∠OPQ = 180° - ∠POQ - ∠PQO

= 180° - 45° - 90°

= 45°.

Therefore, in the △OPQ we have, ∠QOP = ∠OPQ.

Therefore, PQ = OQ = a (say).


OP2 = OQ2 + PQ2

OP2 = a2 + a2

OP2 = 2a2

Therefore,  \(\overline{OP}\) = √2 a (Since, \(\overline{OP}\) is positive)

Therefore, from the right-angled △OPQ we get,

sin  45° = \(\frac{\overline{PQ}}{\overline{OP}} = \frac{a}{\sqrt{2} a} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)

cos  45° = \(\frac{\overline{OQ}}{\overline{OP}} = \frac{a}{\sqrt{2} a} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)

And tan  45° = \(\frac{\overline{PQ}}{\overline{OQ}} = \frac{a}{a} = 1\).

Clearly, csc  45° = \(\frac{1}{sin  45°}\) = √2,

sec  45° = \(\frac{1}{cos  45°}\) = √2

And  cot  45° = \(\frac{1}{tan  45°}\) = 1

Trigonometrical Ratios of 45° are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.

 Trigonometric Functions

11 and 12 Grade Math

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