CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let S=1sin 8+1sin 16+1sin 32++1sin 4096+1sin 8192. If  S=1sin α, where  α(0,90), then  α (in degree) is  


Solution

S=10r=01sin(2r+3)

=10r=0sin(2r+32r+2)sin(2r+2)sin(2r+3)

S=10r=0[cot(2r+2)cot(2r+3)]

S=(cot(22)cot(23))+(cot(23)cot(24))++(cot(212)cot(213))

S=cot 4°cot(8192°)

=cot 4°+tan 2°=cosec(4°)
Here, Every angle is taken in degree.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
View More


People also searched for
View More



footer-image