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Trigonometrical Ratios of 45°

How to find the trigonometrical Ratios of 45°?

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

Trigonometrical Ratios of 45°

Take a point P on OY and draw \(\overline{PQ}
\) perpendicular to 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,  ¯OP = √2 a (Since, ¯OP is positive)

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

sin  45° = ¯PQ¯OP=a2a=12=22

cos  45° = ¯OQ¯OP=a2a=12=22

And tan  45° = ¯PQ¯OQ=aa=1.

Clearly, csc  45° = 1sin45° = √2,

sec  45° = 1cos45° = √2

And  cot  45° = 1tan45° = 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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