How can we prove that` sin^2A+cos^2A =1` ?



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justaguide's profile pic

Posted on (Answer #1)

The identity `sin^2A + cos^2A = 1` has to be proved.

For any angle A it is possible to construct a right triangle ABC such that A is the angle under consideration, AB and BC are the smaller sides and AC is the hypotenuse.

sin A = (opposite side)/(hypotenuse) = `(BC)/(AC)`

cos A = (adjacent side)/(hypotenuse) = `(AB)/(AC)`

`sin^2 A + cos^2A`

= `(BC)^2/(AC)^2 + (AB)^2/(AC)^2`

=> `((BC)^2 + (AB)^2)/(AC)^2`

Applying Pythagoras' Theorem in the right triangle ABC gives `(AB)^2 + (BC)^2 = (AC)^2`

=> `(AC)^2/(AC)^2`

=> 1

This proves that `sin^2 A + cos^2A = 1`

chaobas's profile pic

Posted on (Answer #2)

It can be done through unit circle method. But using pythogorous therorem would be easy.


Let us consider a right angled triangle with one of the angle as "A" which is formed by  "l" as opposite and "b" as adjacent.


Sin A = opposite / hypotenuse 

Cos A = adjacent / hypotenuse


(sinA)^2 + (cos A)^2 = (opposite / hypotenuse )^2 + (adjacent / hypotenuse)^2

= opposite^2+adjacent^2/hypotenuse^2

=hypotenuse^2/hypotenuse^2                                                   [as opposite^2+adjacent^2 = hypotenuse^2 ---pythogorous therorem]




(sin A)^2 + (Cos A)^2 = 1


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