A solution to the equation    is called a zero of the given function f.  As we now show, the zeros of the sine and cosine function are exactly where you might expect them to be.  We have    iff  ,  where n is any integer, and    iff  ,  where n is any integer.

    We show the result for and leave the result for as an exercise.  When we use Identity (5-35),    iff  

            .  

Equating the real and imaginary parts of this equation gives

               and   .  

The real-valued function cosh y is never zero, so the equation    implies that  ,  from which we obtain for any integer n .  Using the values for    in the equation    yields  

            .  

which implies that  ,  so the only zeros for are the values   for  n  an integer.

Exploration.

First, use Mathematica's "Solve" procedure to find some of the solutions to  cos(z) = 0.  

Remark. It is assumed that both  x  and  y  are real numbers. Hence, the only two valid solution in the above list are  .  
This is another way to solve the equation  cos(z) = 0.

We can also list some of the solutions.

Or by showing that the system of equations is satisfied.

Finally, we could just let Mathematica do it.

(c) 2006 John H. Mathews, Russell W. Howell