How do you prove ##(sinx+cosx)^2 = 1+2sinxcosx##?

Use trigonometric identities and the FOIL method.

We are asked to prove that ##(sin x + cos x)^2 = 1 + 2 sin(x) cos(x)##.

1) Change ##(sin x + cos x)^2## to ##(sin x + cos x)(sin x + cos x)## (since the square of any expression is that expression multiplied by itself.)

2) Utilize the FOIL method for multiplying binomials, e.g. ##(sin x + cos x)(sin x + cos x) = (sin x)(sin x) + (sin x)(cos x) + (cos x)(sin x) + (cos x)(cos x)##

3) Simplify and group like terms: ##(sin x)(sin x) + (sin x)(cos x) + (cos x)(sin x) + (cos x)(cos x) = sin^2 x + cos^2 x + 2 sin x cos x##

4) Recall the trigonometric identity which states ##sin^2 x + cos ^2 x =1##, and substitute into (3): ##sin^2 x + cos ^2 x + 2 sin x cos x = 1 + 2 sin x cos x##

5) Use substitution: ##(sin x + cos x)^2 = 1 + 2 sin x cos x##