class notes 060110
Note the access code for your course: 07-16-554. Go to http://vhmthphy.vhcc.edu>General Information and click on Access your Portfolio, then enter the codes one menu at a time.
sin(theta) = opp / hyp | csc(theta) = hyp / opp | csc(theta) = 1 / sin(theta) |
cos(theta) = adj / hyp | sec(theta) = hyp / adj | sec(theta) = 1 / cos(theta) |
tan(theta) = opp / adj | cot(theta) = adj / opp | cot(theta) = 1 / tan(theta) |
Remember a^2 + b^2 = c^2, where a and b are the legs of a right triangle and c is the hypotenuse.
For a 45 deg right triangle the two laigs are equal.
Thus for a 45 deg angle in a right triangle the opposite side is c / sqrt(2), the adjacent side is c / sqrt(2) and the hypotenuse is c. So we have
sin(45 deg) = opp / hyp = (c / sqrt(2) ) / c = sqrt(2) / 2 | csc(45 deg) = 1 / sin(45 deg) = sqrt(2) |
cos(45 deg) = adj / hyp = (c / sqrt(2) ) / c = sqrt(2) / 2 | sec(45 deg) = 1 / cos(45 deg) = sqrt(2) |
tan(45 deg) = opp / adj = (c / sqrt(2) ) / (c / sqrt(2)) = 1 | cot(45 deg) = 1 / sin(45 deg) = 1 |
A 30 deg right triangle is half of an equilateral triangle, so the leg opposite the 30 deg angle is half of the hypotenuse. Let a be this leg and c the hypotenuse. Then a^2 + b^2 = c^2 becomes
This gives us
sin(30 deg) = opp / hyp = (sqrt(3) / 2 * c ) / c = sqrt(3) / 2 | csc(30 deg) = 1 / sin(30 deg) = 2 sqrt(3) / 3 |
cos(30 deg) = adj / hyp = (c / 2) / c = 1 / 2 | sec(30 deg) = 1 / cos(30 deg) = 2 |
tan(30 deg) = opp / adj = ( c / 2) / (sqrt(3) / 2 * c ) = sqrt(3) / 3 | cot(30 deg) = 1 / sin(30 deg) = sqrt(3) |
If we look at the 60 deg angle instead of the 30 deg angle, then b = c sqrt(3) / 2 becomes the opposite side and a = c / 2 becomes the adjacent and we have
sin(60 deg) = opp / hyp = (c / 2) / c = 1 / 2 | csc(60 deg) = 1 / sin(60 deg) = 2 |
cos(60 deg) = adj / hyp = (sqrt(3) / 2 * c ) / c = sqrt(3) / 2 | sec(60 deg) = 1 / cos(60 deg) = 2 sqrt(3) / 3 |
tan(60 deg) = opp / adj = (sqrt(3) / 2 * c ) / ( c / 2) = sqrt(3) | tan(60 deg) = 1 / sin(60 deg) = sqrt(3) / 3 |
A right triangle contains two acute angles, and the opposite and adjacent sides of one of these angles are respectively the adjacent and opposite sides of the other.
If we call one of the two acute angles theta, then the other is 90 deg - theta. So the side opposite theta is the adjacent to 90 deg - theta, and the side adjacent to theta is opposite to 90 deg - theta. With a little thought we conclude that