TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES

For certain specific angles such as 30°, 45° and 60°, which are frequently seen in applications, we can use geometry to determine the trigonometric ratios.

Trigonometric Ratios of 30° and 60°

Let ABC be an equilateral triangle whose sides have length a (see the figure given below). Draw AD perpendicular to BC, then D bisects the side BC.

Then,

BD  =  DC  =  a/2

BAD  =  DAC  =  30°

Now, in right triangle ADB, BAD  =  30° and BD  =  a/2.

In right triangle ADB, by Pythagorean theorem, 

AB2  =  AD2 + BD2

a2  =  AD2 + (a/2)2

a2 - (a2/4)  =  AD2

3a²/4  =  AD2

√(3a2/4)  =  AD

√3 ⋅ a/2  = m AD 

Hence, we can find the trigonometric ratios of angle 30° from the right triangle ADB. 

In right triangle ADB, ABD  =  60°.

So, we can determine the trigonometric ratios of angle 60°. 

Trigonometric Ratio of 45°

If an acute angle of a right triangle is 45°, then the other acute angle is also 45°. 

Thus the triangle is isosceles. Let us consider the triangle ABC with 

B  =  90°

A  =  C  =  45°

Then AB  =  BC.

Let AB  =  BC  =  a.

By Pythagorean theorem,

AC2  =  AB2 + BC2

AC2  =  a2 + a2

AC2  =  2a2

Take square root on each side.

AC  =  a√2

Hence, we can find the trigonometric ratios of angle 45° from the right triangle ABC. 

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

The six trigonometric ratios of angles 0°, 30°, 45°, 60° and 90° are provided in the following table.

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