Author Topic: More Math Problems (Trig Functions)  (Read 5187 times)

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More Math Problems (Trig Functions)
« on: April 03, 2006, 06:07:47 pm »
My MathType trial ran out, so bear with the formating...

Compute Exact Value (Radians, not decimals):
[Is there a way to do these problems without using the unit circle? I don't know how...]

1) arccos(-sqrt(2)/2)
2) arcsin(-1/2)
3) arctan(sqrt(3))
4) arcsin(sin(2pi/3))
5) cos[arccos(-0.7)]
6) csc[arctan(4/3)]

Find The Inverse Function:
y= 4cos(-1/2x)

Write the equation for a sine function with a period of 5pi and a phase shift of pi units to the left and a range of (1,9]

------

a) List the transformations required for y=cos(x) => y=-5cos(4x-2pi)+3 (I put it in the calculator but it's weird...it shrunk alot and stuff)

b) Transform points (using above equation (y=-5cos...) make a table):

(0,1)
(pi/2,0)
(pi,-1)
(3pi/2,0)
(2pi/1)

c) Find the reciprocal equation of y=-5cos(4x-2pi)+3


Find period, domain, range:

1) y=(1/3)cot(2x)
2) 3cos[4(x+pi)] (also find amplitude)
3) y=sec(1/4x)+2

Any help is appreciated, these are due tomorrow morning so if you can do any of them that'd be a great help.

Thanks in advance.

Offline rabbit

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Re: More Math Problems (Trig Functions)
« Reply #1 on: April 03, 2006, 07:14:31 pm »
The inverse is defined as that which is multiplied by the original gives 1 (#4 and #5).

Offline MyndFyre

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Re: More Math Problems (Trig Functions)
« Reply #2 on: April 03, 2006, 09:09:53 pm »
My MathType trial ran out, so bear with the formating...

Compute Exact Value (Radians, not decimals):
[Is there a way to do these problems without using the unit circle? I don't know how...]
Why would you do these w/o the unit circle?
Find The Inverse Function:
y= 4cos(-1/2x)
y/4 = cos(-1/2x) // is that (1/2)x or 1/(2x) ?  assuming former.
arccos(y/4) = -1/2x // you should be able to finish it from here.
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Re: More Math Problems (Trig Functions)
« Reply #3 on: April 03, 2006, 09:23:48 pm »
Yes, the former.

I'd do it w/o unit circle because I haven't memorized it yet..I was supposed to like 2 months ago.

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Re: More Math Problems (Trig Functions)
« Reply #4 on: April 03, 2006, 09:47:20 pm »
I'd do it w/o unit circle because I haven't memorized it yet..I was supposed to like 2 months ago.

Uh yeah do that, it's important and makes things FUCKING EASY.  :)  Thank you, that will be all.
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Re: More Math Problems (Trig Functions)
« Reply #5 on: April 04, 2006, 06:28:01 am »
The inverse is defined as that which is multiplied by the original gives 1 (#4 and #5).

not when the inverse trig function comes first (#4)

Offline Armin

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Re: More Math Problems (Trig Functions)
« Reply #6 on: April 06, 2006, 01:03:03 am »
Yes, the former.

I'd do it w/o unit circle because I haven't memorized it yet..I was supposed to like 2 months ago.
Here's a little pattern I found: For sin(1/2), then measurement in radians will always have 3 as a denominator. Sin[(radical2)/2] will always have a 4 in the denominator, and sin[(radical3)/2] will always have a 6 in the denominator. Cosine is the same thing, but flipped. The cos[(radical3)/2] will have the 3 in the denominator, cos[(radical2)/2] will still have the 4, etc. I had a little trick for remembering the coefficient in front of pi as well, but I forgot it. This is just a little tip to help you out, so meh.
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