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Prove that cos4x=1-8sin^2xcos^2x

Ravindra Srivastava , 13 Years ago
Grade 12th Pass
anser 4 Answers
eaishwary bajpai

you can solve 

cos4x as

cos2(2x) =1-2sin2(2x)

             =1-2(2sinx . cosx)2                          { sin2x = 2sinx. cosx}

             =1-2(4sin2x.cos2x)

            =1-8sin2x.cos2x

LHS=RHS


Last Activity: 13 Years ago
shivang belwariar

cos4x = cos2(2x)

 

2cos^2(2x) - 1

 

2(cos2x)^2 -1

 

2[(2cosx -1 )]^2 -1

 

2[(4cos^2(x)+1-4cox)] -1

 

8cos^2(x) - 8cosx +1

 

approve my answer by clicking yes

Last Activity: 13 Years ago
Priyanka
cos4x=cos2(2x)=cos^2(2x)-sin^2(2x)
            =[cos^2(x)-sin^2(x)]^2-(2sinx cosx)^2
            =[cos^4(x)+sin^4(x)-2cos^(x)sin^2(x)]-[4sin^2(x)cos^2(x)]
            =cos^4(x)+sin^4(x)-6cos^2(x)sin^2(x)
            =[cos^4(x)+sin^4(x)+2cos^2(x)cos^2(x)]-[8cos^2(x)sin^2(x)]
            =[cos^2(x)+sin^2(x)]^2 - 8cos^2(x)sin^2(x)
            = (1)^2 - 8sin^2(x)cos^2(x)
            = 1- 8sin^2(x)cos^2(x)
                                                   HENCE PROOVED
USED FORMULAE
cos2x=cos^2(x)-sin^2(x)
Sin2x=2sinx cosx
(a+b)^2=a^2+b^2+2ab
Last Activity: 6 Years ago
Krish Gupta
You’ll have to go for half angle method :
 
cos4x=cos2(2x)=cos^2(2x)-sin^2(2x)
            =[cos^2(x)-sin^2(x)]^2-(2sinx cosx)^2
            =[cos^4(x)+sin^4(x)-2cos^(x)sin^2(x)]-[4sin^2(x)cos^2(x)]
            =cos^4(x)+sin^4(x)-6cos^2(x)sin^2(x)
            =[cos^4(x)+sin^4(x)+2cos^2(x)cos^2(x)]-[8cos^2(x)sin^2(x)]
            =[cos^2(x)+sin^2(x)]^2 - 8cos^2(x)sin^2(x)
            = (1)^2 - 8sin^2(x)cos^2(x)
            = 1- 8sin^2(x)cos^2(x)
                                                   HENCE PROOVED
USED FORMULAE
cos2x=cos^2(x)-sin^2(x)
Sin2x=2sinx cosx
(a+b)^2=a^2+b^2+2ab
Last Activity: 5 Years ago
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