The value of expression #(2(sin1+sin2+...+sin89))/(2(cos1+cos2+...+cos44)+1)=csctheta, theta in (0,pi/2)# then find #costheta and tantheta#?

Question

7105 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-17T17:59:19

The value of expression #(2(sin1+sin2+...+sin89))/(2(cos1+cos2+...+cos44)+1)=csctheta, theta in (0,pi/2)# then find #costheta and tantheta#

Numerator of LHS

#=2(sin1+sin2+...+sin89)#

#=2(2sin45cos44+2sin45cos43+2sin45cos42+...+2sin45cos1+sin45) #

#=sqrt2(2(cos44+cos43+cos42+...+cos1)+1) #

Now

#(sqrt2(2(cos44+cos43+cos42+...+cos1)+1))/(2(cos1+cos2+...+cos44)+1)=csctheta#

#=>csctheta=sqrt2=csc45#

#=>theta=45#

So #costheta=cos45=1/sqrt2#

And

#tantheta=tan45=1#