Here's the question you clicked on:
Yahoo!
Cos6 . Cos42 . Cos66 . Cos78=?
just trying.. cos 6 = sin 84 = 2sin 42 cos 42....
(78-66) /2 =6 so maybe 2 cos C cos D formula...
(cos 6) (cos 42)(cos 66) (cos 78) = (1/4) * (cos72 + cos60) * (cos120 + cos36) = (1/4) (cos72 + 1/2) * (- 1/2 + cos36) =(1/4)( cos 72 cos36 + 1/2( cos 36- cos72) -1/4 ) use formula for cos a- cos b , = 1/4 ( 2 sin54 sin18 -1/4) =1/4 ( sin36 sin54/ cos 18 -1/4) = 1/4( cos 18 - cos 90) / 2cos18 -1/4) = ans this might seem tedious to read at first, but i believe 've done it correctly.. please reply..
are u sure u typed the qustion properly? isn't there any + or _ sign
Nah..! Question is Correct
i think i got it :) use the formula this : -------------------------------------------- cos(a) . cos(60-a) . cos(60+a) = [cos(3a)]/4 ------------------------------------------- set a = 6 cos(6) . cos(54) . cos(66) = cos(18)/4 so, cos(6) . cos(66) = cos(18)/(4 * cos(54)) .... (1) set a = 18 cos(18) . cos(42) . cos(78) = cos(54)/4 so, cos(42) . cos(78) = cos(54)/(4 . cos(18)) .... (2) multiply both, giving us : cos(6) cos(42) cos(66) cos(78) = 1/16 * cos(18)cos54/(cos(18)*cos(54) = 1/16 * 1 = 1/16
Yup...Thats Correct...thxx.)
even my ans yielded 1/16 ! voila! seems i complicated it too much! :P