How do you evaluate #cos ((15pi)/4)#?

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Answer 1 · 2016-06-25T14:43:21

#cos((15pi)/4) = cos(2*(2pi)-pi/4) = cos(-pi/4) = 1/sqrt(2) = sqrt(2)/2~=0.7071#

We can look at this by considering the angle on the unit circle, where #cos# is the projection on the #x#-axis:

graph{(y^2+x^2-1)((y+0.7071)^2+(x-0.7071)^2-.001)((y)^2+(x-0.7071)^2-.001)=0 [-2.434, 2.433, -1.215, 1.217]}

From this it is easy to see that the same value repeats itself for each full revolution on the circle, i.e. #theta = n*2pi#. In other words:

#cos(n*2pi + theta) = cos(theta)#

In our case we have an angle which is almost 4 full rotations:

#(15pi)/4 = (16pi)/4 - pi/4 = 4pi -pi/4 = 2*(2pi)-pi/4#

Therefore

#cos((15pi)/4) = cos(2*(2pi)-pi/4) = cos(-pi/4) = 1/sqrt(2) = sqrt(2)/2~=0.7071#