ok..
solution:
sin(p/8) = sin(7p/8)
sin(3p/8)=sin(5p/8)
therefore, the equation reduces down to:
2*[sin4(p/8) + sin4(3p/8)]
now, sin2(p/8) = [1-cos(p/4)]/2
also, sin2(3p/8) = [1-cos(3p/4)]/2
i guess now it will be easy to solve the rest
i got this as an extra question.....in my bitsat......
sin4(p/8) + sin4(3p/8) + sin4(5p/8) + sin4(7p/8)
don remembr the options xactly but were lyk
1
2
1/2
3/2
i did post arnd 10 ques ....ppl startd sayin this cheatin so i deletd them ....
plz...post the soln,,,,, i remembr a few more ques ...dun wry i ll post them here....
u plz post the soln...
ok..
solution:
sin(p/8) = sin(7p/8)
sin(3p/8)=sin(5p/8)
therefore, the equation reduces down to:
2*[sin4(p/8) + sin4(3p/8)]
now, sin2(p/8) = [1-cos(p/4)]/2
also, sin2(3p/8) = [1-cos(3p/4)]/2
i guess now it will be easy to solve the rest
yeah but u can substitute values of cos(p/4) and cos(3p/4) now..then square the whole term