Prove the identity of: (sinx - cosx)(sinx + cosx) = 2sin^2 x - 1

Answer 1

(sinx - cosx)(sinx + cosx) = 2sin^2x -1

sin^2x - cos^2x = 2sin^2x -1 subtract sin^2x from both sides to get

-cos^2x = sin^2x -1

add cos^2x + 1 to both sides

1 = sin^2x + cos^2 x

sin^2x + cos^2x = 1

graph a unit circle and sinx= the y coordinate and cosx =the x coordinate

x^2 + y^2 = 1 is the Pythagorean theorem or the Law of Cosines for a right triangle

there are a 100 different ways to prove the Pythagorean Theorem, including one by former President James A. Garfield.

but usually when you get to sin^2x + cos^2x = 1, you stop, as that's a well known trig identity.

or another way:

(sinx + cosx)(sinx - cosx) = 2sinx -1 You can use FOIL to expand the left side or recognize that (a-b)(a+b) = a^2 - b^2 where a= sinx and b= cosx

sin^2x - cos^2x = 2sinx -1 multiply both sides by -1

cos^2x - sin^2x = 1-2sinx

both sides = cos(2x) using the double angle formulas

cos(2x) = cos(2x)

or the sum of angles formula

cos(x+x) = cosxcosx - sinxsinx = cos^2x - sin^2x = 1-2sin^2x

cos2x = cos^2x -sin^2x = 1-sin^2x - sin^2x = 1-2sin^2x