If #sum_(i=1) ^(n)sin(theta_i)=n # then #cos(theta_1)+cos(theta_2)+...+cos(theta_n) =#?

1 Answer
Jan 10, 2018

#cos(theta_1)+cos(theta_2)+...+cos(theta_n)=0#

Explanation:

In the summation #sum_(i=1)^nsin(theta_i)=n#, we have #n# terms and they add up to #n#. However, the maximum possible value of any #sintheta_i# is #1#, when #theta_i=pi/2#.

Hence each term in #sum_(i=1)^nsin(theta_i)=n# is #1# and each #theta_i=pi/2#

therefore each #costheta_i=0# and

#cos(theta_1)+cos(theta_2)+...+cos(theta_n)=0#