I've made you a diagram. (You'll get image posting rights when you've been a member for a while. It's an anti-spammer measure.)

As the point rotates the triangle changes shape. At 90 degrees, the adjacent disappears altogther and the opposite is 1. So you can think of this as showing A/H = zero and O/H = 1 if you like, but there's no triangle because you haven't got three sides.

Have a look at

http://www.mathsisfun.com/geometry/unit-circle.html

About 1/4 of the way down the page is a similar diagram with a 'drag me' point. The triangle degenerates to a line at 90.

and

http://www.mathsisfun.com/algebra/trig- … ircle.html

Bob

]]>Welcome to the forum.

The adjacent/hypotenuse definition only applies to right angled triangles. So strictly speaking you cannot, by that definition, have another angle that is also 90 ... because it wouldn't be a triangle would it.

But you can make a triangle with angles 90, 89, 1 and the cosine of 89 is pretty close to zero as the adjacent is very small.

Similarly 90, 89.9, 0.1 has a cosine that is even closer to zero.

And 90, 89.99, 0.01 is closer still.

.............

Carry on like that and you can see that cosine A tends to zero as A tends to 90.

But you have revealed why a more general definition is needed; one that is the same as the A/H definition for angles between 0 and 90 (exclusive), but also carries on with angles of 90 or over, or even 0 and less. The rotating point definition does exactly that.

Bob

]]>We all know that the cosine of pi/2 is 0 because it is the x coordinate of the point on the unit circle when we move counter clockwise from 0 degree to pi/2. How do you prove that cos of pi/2 equals zero using the law of cosine which is adjacent divided by hypothenuse. I figure since I was able to use the law of sine, law of cosine, and similar triangle to find coordinates of other points on the unit circle, I'd be able to prove that cosine of 90 degree or pi/2 equals to zero as well. However it doesn't look like it is possible. Please reply if you can show how this can be done, thanks.]]>