# SIN COS AND TAN OF SPECIAL ANGLES

## About "Sin cos and tan of special angles"

Sin cos and tan of special angles :

There are two types of special triangles. These triangles are

A 30 - 60 - 90 triangle is a right triangle with a 30° degree angle and a 60° degree angle.

Let "x" be the length of shorter side. The hypotenuse side is twice as long as the shorter side. The longer side is √3 times as long as shorter side.

By using the above special triangle we can find the values of 30 and 60 degrees all six trigonometric ratios.

 sin 30° = x /2x = 1/2cos 30° = (x√3)/(2x)= √3/2tan 30° = x/x√3  = 1/√3sin 60° = (x√3)/(2x)= √3/2cos 60° = x /2x= 1/2tan 60° = x√3/x= √3 cosec 30° = 2x/x=  2sec 30° = (2x)/(x√3)= 2/√3cot 30° = x√3/x = √3/1cosec 60° = (2x)/(x√3)= 2/√3sec 60° = 2x/x= 2cot 60° = x/x√3= 1/√3

Let "x" be the length of equal side. The hypotenuse side is  √2 times the length of either side.

 sin 45° = x /x√2= 1/√2cos 45° = x /x√2= 1/√2tan 45° = x/x = 1 cosec 45° = x√2/x = √2sec 45° = x√2/x= √2cot 45° = x/x = 1

Trigonometric ratios of 0° and 90°

Consider the figure given below which shows a circle of radius 1 unit centered at the origin.

Let P be a point on the circle in the first quadrant with coordinates (x, y).

We drop a perpendicular PQ from P to the x-axis in order to form the right triangle OPQ.

Let <POQ  =  θ, then

sin θ  =  PQ / OP  =  y/1  =  y  (y coordinate of P)

cos θ  =  OQ / OP  =  x/1  =  x  (x coordinate of P)

tan θ  =  PQ / OQ  =  y/x

If OP coincides with OA, then angle θ  =  0°.

Since, the coordinates of A are (1, 0), we have

If OP coincides with OB, then angle θ  =  90°.

Since, the coordinates of B are (0, 1), we have

After having gone through the stuff given above, we hope that the students would have understood "Sin cos and tan of special angles".

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