Why is the cosine of a right angle, 90 degrees, equal to zero?
Your calculation of $\cos (90)$ comes from using a calculator in radians mode, while your angle is in degrees. (You're perfectly correct that the cosine of $90$ degrees is $0$.)
If $\cos A = \sin B$ , does $A+B = 90$ degrees?
This means that if you want to deduce what $\cos (90^\circ)$ should be, you need to make sure that the adjacent side is not just adjacent to the angle $\alpha$, but also to the right angle of the triangle. N.B. Keep in mind that this means that $\cos (90^\circ)$ can only be "observed" in a triangle with two right angles.
I recently learned about the trigonometric functions of angles greater than $90$ degrees, and I'm having a hard time understanding the concept. We worked with a diagram like this: My teacher proce...
let's have a breakdown in the comments made a VERY interesting point that "co-sine" literally means "sin of the complimentary angle" so by definition $\cos A = \sin (90 -A)$.
The angles the trig functions deal with are never the right angle, itself, but the other two. If one of those angles was very close to 90 degrees then the opposite side and the hypotenuse would be nearly equal and very long. The sine of it would be close to 1. The angle can't be 90 degrees: a triangle can't have two right angles.
More explanation - sin and cos are complementary to each other, that's where the name came from - sine and cosine . Complementary angles in a triangle are x and 90-x.