class notes 060110

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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.

If we let x stand for the length of a leg of this triangle, then a = b = x and we have x^2 + x^2 = c^2.
This is the same as 2 x^2 = c^2 or
x^2 = c^2 / 2. The positive square root of both sides gives us
x = sqrt(c^2 / 2) = sqrt(c^2) / sqrt(2) = c / sqrt(2).

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

(c/2)^2 + b^2 = c^2 so that
c^2/4 + b^2 = c^2 and
b^2 = 3/4 * c^2 so that
b = sqrt(3/4 c^2) = sqrt(3/4) sqrt(c^2) = sqrt(3) / sqrt(4) * c = c sqrt(3) / 2.

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

sin(theta) = cos(90 deg - theta)
cos(theta) = sin(90 deg - theta)
tan(theta) = cot(90 deg - theta)
cot(theta) = tan(90 deg - theta)
csc(theta) = secc(90 deg - theta)
sec(theta) = csc(90 deg - theta)