How do you factor and simplify #sin^4x-cos^4x#?

2 Answers
Mia · shr
Aug 29, 2016

#(sinx-cosx)(sinx+cosx)#

Explanation:

Factorizing this algebraic expression is based on this property:

#a^2 - b^2 =(a - b)(a + b)#

Taking #sin^2x =a# and #cos^2x=b# we have :

#sin^4x-cos^4x=(sin^2x)^2-(cos^2x)^2=a^2-b^2#

Applying the above property we have:

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

Applying the same property on#sin^2x-cos^2x#

thus,

#(sin^2x)^2-(cos^2x)^2#
#=(sinx-Cosx)(sinx+cosx)(sin^2x+cos^2x)#

Knowing the Pythagorean identity, #sin^2x+cos^2x=1# we simplify the expression so,

#(sin^2x)^2-(cos^2x)^2#
#=(sinx-Cosx)(sinx+cosx)(sin^2x+cos^2x)#
#=(sinx-cosx)(sinx+cosx)(1)#
#=(sinx-cosx)(sinx+cosx)#

Therefore,
#sin^4x-cos^4x=(sinx-cosx)(sinx+cosx)#

Nghi N
Apr 18, 2017

= - cos 2x

Explanation:

#sin^4x - cos^4 x = (sin^2 x + cos ^2 x)(sin^2 x - cos^2 x) #
Reminder:
#sin^2 x + cos^2 x = 1#, and
#cos^2 x - sin^2 x = cos 2x#
Therefore:
#sin^4x - cos^4 x = - cos 2x#

Related questions
See all questions in Fundamental Identities
Impact of this question
115519 views around the world
You can reuse this answer
Creative Commons License