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

1 Answer
Psykolord1989 .
Jun 20, 2015

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)(sinx+cosx)2=1+2sin(x)cos(x).

1) Change (sin x + cos x)^2(sinx+cosx)2 to (sin x + cos x)(sin x + cos x)(sinx+cosx)(sinx+cosx) (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)(sinx+cosx)(sinx+cosx)=(sinx)(sinx)+(sinx)(cosx)+(cosx)(sinx)+(cosx)(cosx)

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(sinx)(sinx)+(sinx)(cosx)+(cosx)(sinx)+(cosx)(cosx)=sin2x+cos2x+2sinxcosx

4) Recall the trigonometric identity which states sin^2 x + cos ^2 x =1sin2x+cos2x=1, and substitute into (3): sin^2 x + cos ^2 x + 2 sin x cos x = 1 + 2 sin x cos xsin2x+cos2x+2sinxcosx=1+2sinxcosx

5) Use substitution: (sin x + cos x)^2 = 1 + 2 sin x cos x(sinx+cosx)2=1+2sinxcosx

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