Root existence for a function

Question

I want to prove that the following f(x) always has a root. Due to periodicity, it would have many roots if there exist one. Can anybody help me with this? I tried to use intermediate value theorem but I could not come up with the proof.

 

f(x)=\sum_{{i=1 to K}} (n_i/d_i), where

n_i= a_i*sin(x+theta_i)

d_i=2+cos(x-theta_i)

and a_i and theta_i are given for every i. We also know that 0<=a_i<=1 and -pi<=theta_i<=pi .

 

Answer 1

What is the domain of x?

 

As you know, IVT guarantees sine function is zero at 0 and pi.

That will give you the root you want.

 

Will suffice to show that x = -theta along that interval.

 

sin(x+theta) = sinx*cos(theta) - cosx*sin(theta)

 

At theta=0 :   the expression is sin x

At theta=pi/2:   the expression is -cos

At theta=pi: the expression is -sin

 

Sine function is positive in quadrants 1-2, so there is a

nice, clean, and crystal clear sign change

in the angle addition expression for sin(x+thetha) in that interval

 

IVT guarantees a root.

 

end of proof