How to find the trigonometrical Ratios of 45°?
Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anticlockwise sense and starting from the initial position \(\overrightarrow{OX}\) traces out ∠AOB = 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, \(\overline{OP}\) = √2 a (Since, \(\overline{OP}\) is positive)
Therefore, from the rightangled △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.
11 and 12 Grade Math
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