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what should be subtracted from 4(sin^4 30^o+cos^4 60a^o)-2(cos^2 45^o-sin2 60^o) to get 0

Guest May 27, 2017
 #1
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+2

 

 

what a strange question.

 

4(sin^4 30^o+cos^4 60a^o)-2(cos^2 45^o-sin2 60^o)  -    [4(sin^4 30^o+cos^4 60a^o)-2(cos^2 45^o-sin2 60^o)] =0

 

perhaps you want to know what the value is ...

 

\(4(sin^4 30^o+cos^4 60a^o)-2(cos^2 45^o-sin2 60^o)\\ =4(\frac{1}{2^4}+cos^4 60a^o)-2(\frac{1}{2}-sin2 60^o)\\ \text {interpretation of meaning (possible errors)}\\ =4(\frac{1}{2^4}+cos^4 60^o)-2(\frac{1}{2}-sin^2 60^o)\\ =4(\frac{1}{2^4}+\frac{1}{2^4})-2(\frac{1}{2}-\frac{3}{4})\\ =4(\frac{2}{2^4})-2(\frac{-1}{4})\\ =\frac{1}{2}+\frac{1}{2}\\ =1 \)

 

so

4(sin^4 30^o+cos^4 60^o)-2(cos^2 45^o-sin^2 60^o) -1 =0

Melody  May 27, 2017

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