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cos^2a.sin^4a=1/32(cos6a-2cos4a-cos2a+2)

cos^2a.sin^4a=1/32(cos6a-2cos4a-cos2a+2)

Grade:10

1 Answers

Likith Narayan ss
19 Points
9 years ago

Dear   bibek khadka try  this and if satisfied reply me        

   1/32(cos6a-cos2a-2cos4a+2)

let us find the value of  cos6a-cos2a-2cos4a+2                        

          by applying the formula cos c-cos d = 2sin(c+d/2)sin(d-c/2)

     =2sin4a(sin(-2a)-2cos4a+2 

     =-2sin4asin2a-2cos4a+2          

     =2(1-cos4a-sin4asin2a)

     =2(cos^2(2a)+sin^2(2a)-(cos^2(2a)-sin^2(2a))-2sin2acos2a(sin2a) 

    (since cos4a=cos2(2a)=cos^2(2a)+sin^2(2a)) &sin4a=2sin2acos2a & 1=cos^2(2a)+sin^2(2a))

     =2(cos^2(2a) +sin^2(2a) - cos^2(2a) +sin^2(2a)-2sin^2(2a)cos2a)

    =2(2sin^(2a)-2sin^2(2a)cos2a)

     =2(2sin^2(2a)(1-cos2a)      

     =4sin^2(2a)(2sin^2(a))       

     =8sin^2(a)(2sinacosa)^2     

      =8sin^2(a)(4sin^2(a)cos^2(a))

      =32(sin^4(a)cos^2(a)) 

therefore1/32(cos6a-cos2a-2cos4a+2 )=sin^4(a)cos^2(a)

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