(1 + sin x + cos x)^2= 2 (1+sin x)(1+cos x)?

Question

Solve step by step please!

Answer 1

I will use s = sinx, and c= cosx.

(1 + s + c)^2

= (1 + s + c) x (1 + s + c)

= 1 + s + c + s + s^2 + sc + c + cs + c^2

= 1 + 2s + 2c + 2sc + s^2 + c^2

= 1 + 2s + 2c + 2sc + 1

= 2 + 2s + 2c + 2sc

= 2(1 + s + c + sc)

= 2(1 + s + c(1+s))

= 2(1(1+s) + c(1+s))

= 2(1+s)(1+c)

Answer 2

[1 + sin(x) + cos(x)]^2 = 2[1 + sin(x)] [1 + cos(x)]

Start with the more complex side; in this case, it's the right hand side.

LHS = [1 + sin(x) + cos(x)]^2

To make squaring this easier, let's group together the sin(x) and the cos(x), so we can see the FOIL in action.

LHS = [1 + {sin(x) + cos(x)}]^2

FOIL it

LHS = 1 + 2{sin(x) + cos(x)} + {sin(x) + cos(x)}^2

Expand the squared binomial.

LHS = 1 + 2{sin(x) + cos(x)} + {sin^2(x) + 2sin(x)cos(x) + cos^2(x)}

Distribute the 2, and rearrange the terms in the second set of brackets so sin^2(x) and cos^2(x) are together.

LHS = 1 + 2sin(x) + 2cos(x) + sin^2(x) + cos^2(x) + 2sin(x)cos(x)

By the identity, sin^2(x) + cos^2(x) = 1, so

LHS = 1 + 2sin(x) + 2cos(x) + 1 + 2sin(x)cos(x)

LHS = 2 + 2sin(x) + 2cos(x) + 2sin(x)cos(x)

Factor 2 out of everything.

LHS = 2{1 + sin(x) + cos(x) + sin(x)cos(x)}

It may not be obvious, but we can use factor by grouping within the brackets.

LHS = 2{ (1 + sin(x)) + cos(x) (1 + sin(x)) }

Factor (1 + sin(x)) out of the whole thing to get

LHS = 2 (1 + sin(x)) {1 + cos(x)) = RHS