Prove the equality for any real x ( sinx + cosx )^2 = 1 + 2sinx*cosx

To prove the equality means to obtain like expressions
both sides.

We'll start by expanding the square from the
left side. We'll use the formula of expanding the
square:

(a+b)^2 = a^2 + 2ab +
b^2

We'll substitute a and b by sin x and cos x and we'll
get:

( sinx + cosx )^2 = (sin x)^2 + 2sin x*cos x + (cos
x)^2

But the sum (sin x)^2 + (cos x)^2 = 1, from the
fundamental formula of trigonometry.

We'll substitute the
sum by it's value. We'll re-write the expanding of the
square:

( sinx + cosx )^2 = 1 + 2sin x*cos x
q.e.d