9 Oct 2007 06:18:31 IST
It's a good question but it can be solved easily....
Here I don't know how to write theta , so I am using @ for theta
cos2@-cos@=sin4@-sin@
-2 sin(3@/2) sin(@/2) = 2 cos(5@/2) sin(3@/2)
sin(3@/2) { cos(5@/2) + sin(@/2) } =0
Hence sin(3@/2)=0
So sin3@=0 and cos3@=1-(2x0) = 1
Hence the answer is 1
Please rate this .....
Here I don't know how to write theta , so I am using @ for theta
cos2@-cos@=sin4@-sin@
-2 sin(3@/2) sin(@/2) = 2 cos(5@/2) sin(3@/2)
sin(3@/2) { cos(5@/2) + sin(@/2) } =0
Hence sin(3@/2)=0
So sin3@=0 and cos3@=1-(2x0) = 1
Hence the answer is 1
Please rate this .....