Cosine of -90 is 0.
Reciprocal of Cosine is Secant
it would be secant, 1/cosine
No, not necessarily. Cosine theta is equal to 1 only when theta is equal to zero and multiples of 2 pi radians or multiples of 360 degrees. This is because cosine theta is hypotenuse over adjacent, and the ratio 1 only occurs at 0, 360, 720, etc. or 0, 2 pi, 4 pi, etc.
[ cos(Θ) ]-1 = 1/cosine(Θ) = secant(Θ)
Cosine (0) = 1 Sin(0) = 0 The sine and cosine curves are two intertwining curves, that complement each other, hence the words 'Sine (Curve) and Cosine ( COmplementary Curve).
Cosine of -90 is 0.
If the numerator is 0 AND the denominator is not 0, then the ratio is 0. And cosine of 0 is 1.
The Y-Intercept of the cosine function is X = 0, Y = cosine(0) = 1.
Secant is 1 over cosine and cosine 0 equals 1.
Tangent = sine/cosine provided that cosine is non-zero. When cosine is 0, then tangent is undefined.
It is 1.
sin 0 = 0 cos 0 = 1
The inverse of the cosine function is arcosine. The domain is −1 ≤ x ≤ 1 since the range of the cosine function is from -1 to 1. The range is from 0 to pi radians or 0 to 180 degrees.
Zero. Tangent = sine/cosine. sin(0) = 0 and cos(0) = 1, so 0/1 = 0.
int main (void) { puts ("Cosine of 60° is 1/2"); return 0; }
It helps to think as the sine and cosine as coordinates of a unit circle - a circle of radius 1, with center at the origin of the coordinates, i.e., point (0, 0). In this case, as you go around on the circle (starting at the right, coordinates (1, 0), and going counterclockwise), the cosine of the angle is simply the x-coordinate, and the sine of the angle is simply the y-coordinate. At 90°, the x-coordinate is 0, therefore the cosine is 0. Also, at 90° the y-coordinate is 1, therefore the sine is 1 (that's the maximum value it can have).