Flag Trigonometry> multiplesubmultiple angle...
question mark

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

bibek chand khadka , 13 Years ago
Grade 10
anser 1 Answers
Likith Narayan ss

Last Activity: 13 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)

Provide a better Answer & Earn Cool Goodies

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free