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

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

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