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cos^4x - sin^4x = 1-2sin^2x

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help me.. please!

asked Feb 24, 2014 in TRIGONOMETRY by futai Scholar

1 Answer

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Remember the formulae Sin^2x+Cos^2x = 1

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

Left hand side identity = Cos^4x-Sin^4x

= (Cos^2x+Sin^2x)(Cos^2x-Sn^2x)

= 1(Cos^2x-Sin^2x)

= 1-Sin^2x-Sin^2x

= 1-2Sin^2x

= Right hand side identity

 

answered Feb 25, 2014 by ashokavf Scholar

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