# Simplification

• October 11th 2010, 08:40 AM
alexgo
Simplification
I have been trying to figure out if this problem could be simplified more.
$sqrt(cos^2(t) cos(2 sin(t))-sin^2(t))$

Thanks
• October 11th 2010, 08:54 AM
Jhevon
Quote:

Originally Posted by alexgo
I have been trying to figure out if this problem could be simplified more.
$sqrt(cos^2(t) cos(2 sin(t))-sin^2(t))$

Thanks

as in $\displaystyle \sqrt{\cos^2 t \cdot \cos (2 \sin t) - \sin^2 t}$ ? Is this right?
• October 11th 2010, 08:55 AM
alexgo
Quote:

Originally Posted by Jhevon
as in $\displaystyle \sqrt{\cos^2 t \cdot \cos (2 \sin t) - \sin^2 t}$ ? Is this right?

yup
• October 11th 2010, 08:58 AM
Jhevon
Quote:

Originally Posted by alexgo
yup

well, not really. i don't see any simplification that would be worth the effort.

why do you want to do this? what was the original problem?
• October 11th 2010, 09:04 AM
alexgo
Quote:

Originally Posted by Jhevon
well, not really. i don't see any simplification that would be worth the effort.

why do you want to do this? what was the original problem?

Original Problem is actually a Calc 3 problem. I was looking for Tangent vector of $r(t)=sin(sint) i + cos(sint) j + Cost k$
• October 11th 2010, 09:11 AM
Jhevon
Quote:

Originally Posted by alexgo
Original Problem is actually a Calc 3 problem. I was looking for Tangent vector of $r(t)=sin(sint) i + cos(sint) j + Cost k$

oh! you mean the unit tangent vector? i suppose you were trying to simplify |r'(t)|

in that case, you're way off. your |r'(t)| is incorrect. for the correct one, it is 1.

Hint: you made a sign error. there is a minus (at least one) that should be a plus in what you did.
• October 11th 2010, 09:34 AM
alexgo
Quote:

Originally Posted by Jhevon
oh! you mean the unit tangent vector? i suppose you were trying to simplify |r'(t)|

in that case, you're way off. your |r'(t)| is incorrect. for the correct one, it is 1.

Hint: you made a sign error. there is a minus that should be a plus in what you did.

AHHH, so when I square I take the negative with it, right?
for $r'(t)=cost \cdot cos(sint) i - cost \cdot sin(sint) j- sint k$
for $|r'(t)|=\sqrt{(cost \cdot cos(sint))^2 +(-cost \cdot sin(sint))^2 +(-sint)^2}$
$|r'(t)|=\sqrt{cos^2t \cdot (cos^2(sint) + sin^2(sint)) +sin^2t}$
$|r'(t)|=\sqrt{cos^2t \cdot (1) +sin^2t}$
$|r'(t)|=1$
• October 11th 2010, 10:08 AM
Jhevon
Quote:

Originally Posted by alexgo
AHHH, so when I square I take the negative with it, right?
for $r'(t)=cost \cdot cos(sint) i - cost \cdot sin(sint) j- sint k$
for $|r'(t)|=\sqrt{(cost \cdot cos(sint))^2 +(-cost \cdot sin(sint))^2 +(-sint)^2}$
$|r'(t)|=\sqrt{cos^2t \cdot (cos^2(sint) + sin^2(sint)) +sin^2t}$
$|r'(t)|=\sqrt{cos^2t \cdot (1) +sin^2t}$
$|r'(t)|=1$

(Yes)