Pythagorian Identities





 

Let “OPQ” be a triangle where angle POQ is \theta , and it’s base be “x” and perpendicular “y” as shown in the picture below:

Pythagorian Identities

Pythagorian Identities

 

Then,

Appealing to the Pythagorian theorem, we have:

OQ^2 + QP^2 = OP^2

Now let us suppose op be “r” then:

x^2 + y^2 = r^2

And as:

x = r \cos \theta and y = r \sin \theta

 

We have:

r^2 ( \cos ^2 \theta + \sin ^2 \theta = r^2

Thus: \sin ^2 \theta + \cos ^2 \theta = 1

This is called the fundamental Pythagorian identity of trigonometry.

From this we can also develop other identities as:

 

Dividing both side of fundamental Pythagorian identity by  \cos ^2 \theta

\frac{\sin ^2 \theta}{\cos ^2 \theta} + \frac{\cos ^2 \theta}{\cos ^2 \theta} = \frac{1}{\cos ^2 \theta}

 

Hence ,1 + \tan ^2 \theta = \sec ^2 \theta

 

And now dividing both side of fundamental Pythagorian identity by  \sin ^2 \theta

\frac{\sin ^2 \theta}{\sin ^2 \theta} + \frac{\cos ^2 \theta}{\sin ^2 \theta} = \frac{1}{\sin ^2 \theta}

 

Thus , 1 + \cot ^2 \theta = \csc ^2 \theta



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