It doesn't exist. The maximum value of the cosine is 1.00, so no angle can have a cosine of (pi), because (pi) is more than 3.
udefined
The cosine of 2pi is 1. In fact, for every integer N, the cosine of 2 N pi is 1.
The sine curve is exactly the same as the cosine curve shifted pi/2 radians to the left
0.5877852523
It doesn't exist. The maximum value of the cosine is 1.00, so no angle can have a cosine of (pi), because (pi) is more than 3.
0.99847149863
udefined
The cosine of 2pi is 1. In fact, for every integer N, the cosine of 2 N pi is 1.
A "zero of a function" is a point where the dependent value (usually, Y) is zero. In the function f(x) = x2 - 2, for example, there are zeroes at -1.414 and +1.414.The zeroes of the sine function are at all integer multiples of pi, i.e. 0, pi, 2pi, 3pi, etc. The zeroes of the cosine function are at the same points plus pi/2, i.e. pi/2, 3pi/2, 5pi/2, etc.Another way to look at this is that the zeroes of sine are the even multiples of pi/2, and the zeros of cosine are the odd multiples of pi/2.
This is going to require some visualization. Cosine is defined as the x-value on the unit circle. If you picture where a point would be, for example, at the angle of pi/6 (30°) you get a coordinate of (√(3)/2 , 1/2) so cosine is √(3)/2 and sine is 1/2 To find a negative angle you take the reflection across the x-axis. Since this does not chance the x-value, only the y, cosine does not change. The coordinates of -(pi/6) (-30°) are (√(3)/2 , -1/2). cos(-x) = cos(x) sin(-x) = - sin(x)■
The sine curve is exactly the same as the cosine curve shifted pi/2 radians to the left
Yes, except at odd multiples of pi/2 radians, where the cosine is zero so that the division is not defined.
at a 45 degree angle, or pi/4
0.75
Yes they are. Both have a a period of 2 pi
0.5877852523