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

1 Answer
Jun 20, 2015

Answer:

Use trigonometric identities and the FOIL method.

Explanation:

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#