How to prove sin^6x+cos^6x=1-3sin^2xcos^2x?

3 Answers
May 23, 2018

LHS=sin^6x+cos^6x

=(sin^2x)^3+(cos^2x)^3

Using formula

a^3+b^3=(a+b)^3-3ab(a+b)

=(sin^2x+cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x)

=1^3+3sin^2xcos^2x*1

=1-3sin^2xcos^2x=RHS

May 23, 2018

See below.

Explanation:

Proof

sin^6(x)+cos^6(x)

=(sin^2(x)+cos^2(x))(sin^4(x)-sin^2(x)cos^2(x)+cos^4(x)) (1)

=sin^4(x)-sin^2(x)cos^2(x)+cos^4(x) (2)

=(1-2sin^2(x)cos^2(x))-sin^2(x)cos^2(x) (3)

=1-3sin^2(x)cos^2(x)

More detailed explanations

(1) since a^3+b^3=(a+b)(a^2-ab+b^2),

(2) since sin^2(x)+cos^2(x)=1,

(3) since (sin^2(x)+cos^2(x))^2=1<=>sin^4(x)+cos^4(x)+2sin^2(x)cos^2(x)<=>sin^4(x)+cos^4(x)=1-2sin^2(x)cos^2(x)

May 23, 2018

Call sin^2 x = u, and cos^2 x = v
We get, knowing that (u + v = 1):
u^3 + v^3 = (u + v)(u^2 - uv + v^2) = u^2 + v^2 - uv (1)
Note that:
u^2 + v^2 = (u +v)^2 - 2uv = 1 - 2uv
Equation (1) becomes:
u^3 + v^3 = 1 - 2uv - uv = 1 - 3uv

sin^6 x + cos^6 x = 1 - 3sin^2 x.cos^2 x