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Lunare
05-14-2008, 10:34 PM
Hi. I seem to have some trouble when an equation with tan surfaces. If someone could help me it would be great. want to know if i actually even started correctly.

1) suppose tan x = -4/3 with 0 < x < pi. Find the exact value of tan (x + pi)

If i start this with the sum/difference identity of tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi)

Then this should get me through to the answer right?
I hope.

2) suppose tan x = -4/3 with 0 < x < pi. Find the exact value of cot x.

Ok for this one I chose a double angle identity but the more i looked at my steps and solution the more i doubted it. Suffice to say i erased that answer.

Any help would be appreciated.
Thank you
Lunare

Mrspi
05-14-2008, 11:04 PM
Hi. I seem to have some trouble when an equation with tan surfaces. If someone could help me it would be great. want to know if i actually even started correctly.

1) suppose tan x = -4/3 with 0 < x < pi. Find the exact value of tan (x + pi)

If i start this with the sum/difference identity of tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi)

Then this should get me through to the answer right? RIGHT....and tan pi = 0. So, you will get tan (x + pi) = tan x
I hope.


2) suppose tan x = -4/3 with 0 < x < pi. Find the exact value of cot x.

Ok for this one I chose a double angle identity but the more i looked at my steps and solution the more i doubted it. Suffice to say i erased that answer.

You may be making this too complicated. Remember that cot x = 1 / tan x. If tan x = -4/3, then what is cot x?


Any help would be appreciated.
Thank you
Lunare

Loren
05-14-2008, 11:42 PM
Aren't you allowed to draw a sketch? If you were to draw a sketch and label the abscissa and ordinate you would end up with two right triangles, one in quadrant II and one in quadrant IV. The answers becomes quite obvious.
If you must use identities, you can easily check your result with the sketch.

<If i start this with the sum/difference identity of tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi).

tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi) means \tan(x+\pi) = \tan x + \frac{\tan \pi}{1}-\tan x\cdot \tan \pi
Maybe you mean tan (x + pi) = (tan(x) + tan (pi)) /( 1 - tan(x) tan (pi))
It seems like the long way around to me.

Lunare
05-15-2008, 01:55 AM
Hi. I seem to have some trouble when an equation with tan surfaces. If someone could help me it would be great. want to know if i actually even started correctly.

1) suppose tan x = -4/3 with 0 < x < pi. Find the exact value of tan (x + pi)

If i start this with the sum/difference identity of tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi)

Then this should get me through to the answer right? RIGHT....and tan pi = 0. So, you will get tan (x + pi) = tan x
I hope.

Yes. That was my solution. I just didn't think it was correct.


2) suppose tan x = -4/3 with 0 < x < pi. Find the exact value of cot x.

Ok for this one I chose a double angle identity but the more i looked at my steps and solution the more i doubted it. Suffice to say i erased that answer.

You may be making this too complicated. Remember that cot x = 1 / tan x. If tan x = -4/3, then what is cot x?
I might be. I just didn't think that it was that easy of an answer especially when we are in the section of sum/difference, half-angle and double angle identities.


Thank you Mrspi

Any help would be appreciated.
Thank you
Lunare

Lunare
05-15-2008, 02:01 AM
Aren't you allowed to draw a sketch? If you were to draw a sketch and label the abscissa and ordinate you would end up with two right triangles, one in quadrant II and one in quadrant IV. The answers becomes quite obvious.
If you must use identities, you can easily check your result with the sketch.

<If i start this with the sum/difference identity of tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi).

tan (x + pi) = tan(x) + tan (pi) / 1 - tan(x) tan (pi) means \tan(x+\pi) = \tan x + \frac{\tan \pi}{1}-\tan x\cdot \tan \pi You're right. I apologize.
Maybe you mean tan (x + pi) = (tan(x) + tan (pi)) /( 1 - tan(x) tan (pi)) Yes this is the one I meant.
It seems like the long way around to me.
Yes, but this teacher is fickle. We can't even skip simple common steps or he deducts points whether on tests or homework. So i do what i can.


Thank you Loren