# Homework Help: Normal of trigonometry points

1. Nov 5, 2009

### hanelliot

1. The problem statement, all variables and given/known data
x = (-4$$\pi$$sin$$\pi$$t, 4$$\pi$$cos$$\pi$$t, 1)
what is the length of x?

2. Relevant equations

3. The attempt at a solution
Well, I do realize that length of something is calculated by sqrt(x^2 + y^2) etc.
However, when I plug in the numbers above, I get sqrt(16$$\pi$$^3 + 1), but the answer is sqrt[1 + (4$$\pi$$)^2)]. If anyone can show me step by step (trig is my weak point), I'd really appreciate it. I'm guessing (sin^2$$\pi$$t + cos^2$$\pi$$t) = 1, instead of $$\pi$$? If so, why?

2. Nov 5, 2009

### HallsofIvy

You should have learned long ago that $sin^2(x)+ cos^2(x)= 1$ for all x, even if $x= \pi t$!

3. Nov 5, 2009

### Staff: Mentor

Yes, sin2A + cos2A = 1 for any A. This is the only trig in this problem. The rest is algebra, which also seems to be a weak point.

How did you get sqrt(16$\pi$3 + 1)? Show what you did to get this and we can set you straight.

4. Nov 5, 2009

### HallsofIvy

I do hope you are NOT arguing that you can take a "$\pi$ out of the function and saying that:

"$sin^2(\pi t)+ cos^2(\pi t)= (\pi sin^2(t))^2+ (\pi cos^2(t))^2$$= \pi^2$".

sin(at) is NOT equal to a sin(t) and cos(at) is NOT equal to a cos(t)!

5. Nov 5, 2009

### hanelliot

lol yeah, been away from math for a long time so did forget it. don't worry about 16pi^3, it was done quickly and obv wrong.