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

3 Answers
P dilip_k
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#

e52fa787
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)#

Nghi N
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#

Related questions
Impact of this question
42654 views around the world
You can reuse this answer
Creative Commons License