Question 811471
Using the trig identity cos(2x) = 2cos^2(x) - 1 we can write the LHS of the equation as
2cos^2(x) - 1 + 2sin^2(x) -> 2(cos^2(x) + sin^2(x)) - 1
Since sin^2(x) + cos^2(x) = 1, the LHS becomes 2 - 1 = 1