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

Answer 1

Zoe Austin

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

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)#

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Answer 2

Elijah Alexander

= - cos 2x

#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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.