Sin^2x+Cos^2x=1 How to prove it?

Question

Tks

Answer 1

Using Ó¨ as angle and x , y and r in the usual convention :-

sin Ó¨ = y / r

cos Ó¨ = x / r

sin ² Ө = y² / r²

cos ² Ө = x² / r²

sin ² Ө + cos ² Ө = y ² / r ² + x ² / r ²

sin ² Ө + cos ² Ө = ( x ² + y ² ) / r ²

sin ² Ө + cos ² Ө = r ² / r ²

sin ² Ө + cos ² Ө = 1

Answer 2

Consider a right triangle with lengths a, b, and c, where c is the hypotenuse.

(sinx)^2 + (cos x)^2 = (a/c)^2 + (b/c)^2 = (a^2 + b^2) / c^2.  But  by the Pythagorean Theorem this equals c^2 / c^2 = 1.

Answer 3

Imagine a rt.< triangle with the hypotenus=1,then

select an interior acute angle x. Since

sinx= opposit side/1---(1)

cosx=adjacent side/1---(2)

sin^2x+cos^2x=

(opposit side)^2+(adjacent side)^2

=1---- [(1)^2+(2)^2]