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Prove that `cos^4A - sin^4A +1 = 2cos^2A`

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user1504207 | Student, Grade 11 | eNotes Newbie

Posted April 9, 2013 at 11:00 AM via web

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Prove that `cos^4A - sin^4A +1 = 2cos^2A`

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mjripalda | High School Teacher | (Level 3) Educator

Posted April 9, 2013 at 11:17 AM (Answer #1)

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`cos^4A-sin^4A+1=2cos^2A`

First, group the first two terms and factor it.

`(cos^4A-sin^4A)+1=2cos^2A`

`(cos^2A-sin^2A)(cos^2A+sin^2A)+1=2cos^2A`

Then, apply the Pythagorean identity which is  `cos^2theta + sin^2theta=1` .

 

`(cos^2A-sin^2A)(1)+1=2cos^2A`

`cos^2A-sin^2A + 1=2cos^2A`

Then, group the second and last term at the left side of the equation.

`cos^2A+(-sin^2A+1)=2cos^2A`

`cos^2A+(1-sin^2A)=2cos^2A`

To simplify the expression inside the parenthesis, apply the Pythagorean identity again.

`cos^2A+cos^2A=2cos^2A`

`2cos^2A=2cos^2A`

Since left side simplifies to `2cos^2A` which is the same term with the right side, hence it proves that the given equation is an identity.

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oldnick | Valedictorian

Posted April 10, 2013 at 12:09 AM (Answer #2)

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cos^4 A - sin^4 A +1 = (cos^2 A + sin^2 A)(cos^2 A - sin^2 A) +1 since sin^2 A + cos^2 A = 1 we get: cos^2 A - sen^2 A + 1 = cos^2 A -( 1-cos^2 A )+1 = cos^2 A -1+ cos^2 A +1 = 2cos^2 A
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user1450001 | Student, Undergraduate | Valedictorian

Posted April 10, 2013 at 11:45 AM (Answer #3)

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Let x=sin(A)^2 and y=cos(A)^2, then

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

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