# Homework Help: Simplifying Trig Identities

1. Jun 1, 2014

### mill

1. The problem statement, all variables and given/known data

The length of the curve r(t) = cos^3(t)j+sin^3(t)k, 0 =< t <= pi/2
is

2. Relevant equations

AL in polar = ∫sqrt(r^2 + [dr/dθ]^2)

3. The attempt at a solution

I am having trouble simplifying the terms within the square root. What method should I use to deal with the pieces?

r^2 = (cos^3(t))^2 + 2 cost^3(t)sin^3(t) + (sin^3(t))^2

dr/dθ = 3(cos(t)^2)sin(t) + 3(sin(t)^2)cos(t)

[dr/dθ]^2 = 9[cos^4(t)sin^2(t) + 2cos^3(t)sin^3(t) + sin^4(t)cos^2(t))

sqrt(cos^3(t))^2 + 2 cost^3(t)sin^3(t) + (sin^3(t))^2 + 9[cos^4(t)sin^2(t) + 2cos^3(t)sin^3(t) + sin^4(t)cos^2(t))

Simplified a bit:

sqrt(1 + 4cos^3(t)sin^3(t) + 9 cos^4(t)sin^4(t) +9sin^4(t)cos^2(t))

How would I further simplify from this?

2. Jun 1, 2014

### haruspex

There seem to be some inconsistencies in your notation.
From the equation for r(t), I guess j and k are unit vectors, so r(t) is a vector. But your equation for arc length treats r as a scalar. This seems to have led to an error here:
$r^2 = (\cos^3(t))^2 + 2 \cos^3(t)\sin^3(t) + (\sin^3(t))^2$

3. Jun 1, 2014

### mill

Oh, I see. Thanks.