Expert: Ahmed Salami - 8/26/2009

Question
how do I proceed?

tan(-6pi) + cos(9pi/4)

Answer
Hi Kamila,
First, some useful things to remember
sin(-x) = -sinx
cos(-x) = cosx
tan(-x) = -tanx
and in trigonometry when an angle x is greater than 2π,
sinx = sin(x - 2πm)
where m is an integer.
The same applies as well to both the cosines and tangents of such angles. What this means is that when we have an angle greater than 2π we can remove any integral multiple of 2π until we get an angle between 0 and 2π so that we can easily find what we need using our normal methods.
Now,
tan(-6π) = - tan(6π)
           = - tan(6π - 6π)
           = - tan(0)
           = 0
Notice here that we have taken m = 3
cos(9π/4) = cos(9π/4 - 2π)
         = cos(π/4)
         = 1/√2
         = √2/2
And so,
tan(-6π) + cos(9π/4) = 0 + √2/2
                    = √2/2

Regards