Some math questions- PreCalculus

[aghtar]

Hi guys, well I have some problems I'm struggling with.

Rewrite sqrt((1-cos(10x))/12) using a half angle identity..

So I used the sin(x/2)= +/- sqrt((1-cos(x))/2) identity, and came up with this:

sin (10x/12) = +/- sqrt(1-cos(10x))/12) is that correct?

My other question concerns rectangular and polar coordinates..

If the professor gave us this instructions:
Provide and equivalent point in the different coordinate system. (There may be more than one correct answer, but you only need provide one).

Then gave us this chart:

Rectangular	Polar
(exact values)	(-2, 3 /4)
(-3, -4)	(round to three decimal places)

--Note --The spaces above the statements within parentheses should be where we write our answers.

How should I go about that problem? I missed that class, and have no idea how to begin.

Thanks for any help.

 

[srmichael]

Hi guys, well I have some problems I'm struggling with.

Rewrite sqrt((1-cos(10x))/12) using a half angle identity..

So I used the sin(x/2)= +/- sqrt((1-cos(x))/2) identity, and came up with this:

sin (10x/12) = +/- sqrt(1-cos(10x))/12) is that correct?

My other question concerns rectangular and polar coordinates..

If the professor gave us this instructions:
Provide and equivalent point in the different coordinate system. (There may be more than one correct answer, but you only need provide one).

Then gave us this chart:

Rectangular	Polar
(exact values)	(-2, 3 /4)
(-3, -4)	(round to three decimal places)

--Note --The spaces above the statements within parentheses should be where we write our answers.

How should I go about that problem? I missed that class, and have no idea how to begin.

Thanks for any help.

First, only one question oer thread, please. I'll let you slide this time since you are new

For the first one, I'm not sure I understand your first equation, but \(\displaystyle \displaystyle\displaystyle\sin(\frac{10x}{12})=\pm\sqrt{\frac{1-\cos(\frac{10x}{6})}{2}}\)

For the second one, remember that given r and theta, \(\displaystyle \displaystyle x=r\cdot\cos(\theta)\) and \(\displaystyle \displaystyle y=r\cdot\sin(\theta)\)

And given x and y, \(\displaystyle \displaystyle r=\sqrt{x^2+y^2}\) and \(\displaystyle \displaystyle \theta=\arctan{\frac{y}{x}} + \pi\) if x < 0

Try these hints out and see what you get.