Can you help me prove the following trig equation? (Sin^2 X) + (Cos^2 X) = 1

Answer 1

In the unit circle

sin² x = y²

cos² = x²

x² + y² = 1

sin² x + cos² x = 1

Answer 2

Given a right triangle

sin²x + cos²x = 1

sin x = opp/hyp = AB/AC

cos x = adj/hyp = BC/AC

sin²x = (AB/AC)²

cos²x = (BC/AC)²

sin²x + cos²x = 1

AB²/AC² + BC²/AC²

(AB²+BC²)/AC²

using the pythagorean theorum (a² + b² = c² )

AB² + BC² = AC²

substitute the numerator with AC²

AC²/AC² = 1

Answer 3

start with the pythagorean theorem and an angle x

call opposite side o and adjacent side a

and hypotenuse h

Then o^2 + a^2 = h^2

now divide both sides by h^2 which gives us

o^2/h^2. + a^2/h^2 = h^2/h^2

o^2/h^2 is the same as sin^2

a^2/h^2 is the same as cos^2 and

h^2/h 2 is 1 so

sin^2 + cos^2 = 1