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).

Now,

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

From Trigonometrical Ratios of 45° to HOME PAGE


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.