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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

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