Integral Calculus, More Trig Functions

More Trig Functions

With sine and cosine defined for all real numbers, we can now define the other 4 trig functions.

Tangent, written tan(x), is sin(x)/cos(x).
Cotangent, written cot(x), is cos(x)/sin(x).
Secant, written sec(x), is 1/cos(x).
Cosecant, written cosec(x), is 1/sin(x).

If you have trouble remembering what goes with what, there is one "co" for each reciprocal pair. Tangent and cotangent are reciprocal, sin and cosecant are reciprocal, and cosine and secant are reciprocal.

Compute their derivatives using the quotient rule. I include sine and cosine for completeness.

sin(x) → cos(x)
cos(x) → -sin(x)
tan(x) → sec²(x)
cot(x) → -cosec²(x)
sec(x) → sec(x)tan(x)
cosec(x) → -cosec(x)cot(x)

The arc functions are the inverse functions. They are usually indicated by a leading a. Thus atan(tan(x)) = x, and asec(sec(x)) = x, and so on.

Use the inverse function rule to find the derivatives of these arc trig functions. For instance, if y = atan(x) then x = tan(y). The derivative of x with respect to y is the derivative of tangent, or secant squared. Therefore the derivative of y with respect to x is the reciprocal, or 1 over sec²(y). But how do we get back to x? We know that x is the tangent of y, so think of y as an angle, θ if you prefer. Draw a triangle with base 1 and height x. Sure enough, x = tan(θ). The second is the hypotenuse, which equals sqrt(x²+1) (pythagorean theorem), divided by the base, which equals 1. Thus secant squared = 1+x², and the derivative of atan(x) is 1 over 1+x². Similar reasoning gives formulas for the other derivatives. Here they are in one table.

asin(x) → 1 / sqrt(1-x²)
acos(x) → -1 / sqrt(1-x²)
atan(x) → 1 / (1+x²)
acot(x) → -1 / (1+x²)
asec(x) → 1 / (x×sqrt(x²-1))
acosec(x) → -1 / (x×sqrt(x²-1))

You've noticed that the derivatives clump together in ± pairs. Given any x, asin(x) gives one angle and acos(x) gives the other. These are complementary angles in a right triangle. Thus asin(x) + acos(x) is always 90°. Take derivatives and the right side drops to 0, hence the derivative of arcsine must be the opposite of the derivative of arccosine. Similar reasoning holds for tangent and cotanget, secant and cosecant.

The Ladder in the Hallway

Let's practice some trig differentiation. Two hallways meet at a right angle. They have widths a and b. You are carrying a long ladder on its side, and you need to turn the corner. What is the longest ladder, in terms of a and b, that will make the turn?

Pivot the ladder around the inner corner of the hallways. The shortest line corresponds to the longest ladder that will make the turn. If the horizontal hallway has width a, and the line is at angle θ to the horizontal, its length is b/c + a/s, where c and s are the cosine and sine of θ respectively. Differentiate, and set to 0, giving bs³ = ac³. The tangent of θ is the cube root of a/b. Call this expression w. Draw a right triangle whos base is 1 and altitude is w. (This gives the correct tangent.) The pythagorean theorem gives the hypotenuse, and the other trig ratios. The ladder has the following length.

b/c + a/s

(b + a/w) × sqrt(w²+1)

b^1/3 × (b^2/3 + a^2/3) × sqrt(w²+1)

(a^2/3 + b^2/3) × sqrt(a^2/3 + b^2/3)

(a^2/3 + b^2/3)^3/2

The last formulation is symmetric in a and b, as one would expect. Set a = b and get 2×sqrt(2)×b, as one would expect. Finally, let a approach 0, and the length approaches b, as one would expect.