Question

50 points !!! Value of : Sin⁴π\8 + Sin⁴3π\8 + Sin⁴5π\8 + Sin⁴7π\8

Answer 1

it's easy

             sin⁴π/8 +   sin⁴3π/8   +  sin⁴5π /8    + sin⁴7π/8

   = sin⁴ π/8     +   sin⁴(π - 5π/ 8)    +  sin⁴5π/8   +sin⁴(π - π/8)

   =  sin⁴ π/8  +sin⁴ 5π /8    +sin⁴5 π /8  + sin⁴π/8

           =        2sin⁴ π/8   +2sin⁴5π/8

     =  2(sin⁴  π/8  + sin⁴5π/8)

       =2(sin⁴  π /8 +sin⁴(π/2 + π/8) )

     =  2(sin⁴  π /8 +cos⁴π/8) = 2 ( 1 - 2sin²π/8cos²π/8 )  = 2(1 - sin²2π/8 /2)

         = 2 (1 - sin²π/4 / 2 )  =  2(1- 1/4)  = 3/2

hope u'll get ........I've used here some formulas  ...........if doubt then ask

Answer 2

hi friend,


sin⁴π/8 +sin⁴3π/8+(sin(π-3π/8))⁴ +(sin(π-π/8)⁴


=sin⁴π/8 +sin⁴3π/8 +sin⁴π/8 +sin⁴3π/8

=2(sin⁴π/8 +sin⁴3π/8 )

=2(sin⁴π/8+(sin(π/2-π/8))⁴)

=2(sin⁴π/8 +cos⁴π/8)-----(1)


we know that cos²x+sin²x=1

squaring on both sides

cos⁴x+sin⁴x+2sin²xcos²x=1

cos⁴x+sin⁴x=1-2sin²xcos²x-----(2) by taking x=π/8


by substituting (2)in (1)

=2(1-2sin²xcos²x)

=2(1-2sin²π/8cos²π/8)

=2(1-sin²(2π/8)/2)

=2(1-1/4)

=2(3/4)

=3/2


I hope this will help u :)