Where do I start this triq problem?

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June 17th 2008, 03:32 PM
Darkhrse99

Where do I start this triq problem?

I need to find the 5 triq. expressions of $csc=-4$ I know the properties, but I'm not sure which expression I need to start with to find the other 4 expressions. $csc=\frac{1} {sin}$ is sin the receprical of-4, so it will be $sin= \frac{1}{4}$ ?

Thanks
Jason
June 17th 2008, 04:02 PM
Jhevon

Quote:

Originally Posted by Darkhrse99

I need to find the 5 triq. expressions of $csc=-4$ I know the properties, but I'm not sure which expression I need to start with to find the other 4 expressions. $csc=\frac{1} {sin}$ is sin the receprical of-4, so it will be $sin= \frac{1}{4}$ ?

Thanks
Jason

I ignored this in your last thread, but i must say something now. you cannot have trig functions standing alone. writing csc and sin makes no sense! write csc(x) and sin(x) or whatever variable you want, they must be operating on something. that being said, you are right. csc(x) = 1/sin(x)

so if csc(x) = -4, then sin(x) = -1/4

now continue with the others
June 17th 2008, 04:45 PM
Darkhrse99

Sorry about that. I'll make sure to write it as Tan (x) as that is the variable we are using.
June 17th 2008, 05:12 PM
Darkhrse99

Does $Cos(x) = \frac{\sqrt17} {4}$ and $tan (x) =\frac{1} {\sqrt17}$ ?
June 17th 2008, 05:39 PM
Jhevon

Quote:

Originally Posted by Darkhrse99

Does $Cos(x) = \frac{\sqrt17} {4}$ and $tan (x) =\frac{1} {\sqrt17}$ ?

Nope. where did 17 come from?
June 17th 2008, 05:48 PM
Darkhrse99

Sin (x)was $\frac{1} {4}$ so $1^2+ 4^2=17, x={\sqrt17}$ ? Adjacent = ${\sqrt17}$
June 17th 2008, 05:51 PM
Jhevon

Quote:

Originally Posted by Darkhrse99

Sin (x)was $\frac{1} {4}$ so $1^2+ 4^2=17, x={\sqrt17}$ ?

no. 4 is the length of the hypotenuse. remember, sine = opposite/hypotenuse. thus you want 4^2 - 1^2.

alternatively, you can use the famous trig identity: $\sin^2 x + \cos^2 x = 1$
June 17th 2008, 05:55 PM
Darkhrse99

so it is $cos (x)=\frac{\sqrt15} {4}$ and $tan (x)= \frac{1}{\sqrt15}$ ?
June 17th 2008, 06:00 PM
Jhevon

Quote:

Originally Posted by Darkhrse99

so it is $cos (x)=\frac{\sqrt15} {4}$ and $tan (x)= \frac{1}{\sqrt15}$ ?

we need to know what quadrant we are in to know what the sign of the cosine (and hence the tangent) is
June 17th 2008, 06:11 PM
Darkhrse99

CSC (t) is in the 4th quad, so $cos(x)=\frac{\sqrt15} {4} and Tan (x)=\frac{-1} {\sqrt15}$
June 17th 2008, 06:15 PM
Jhevon

Quote:

Originally Posted by Darkhrse99

CSC (t) is in the 4th quad, so $cos(x)=\frac{\sqrt15} {4} and Tan (x)=\frac{-1} {\sqrt15}$

correct (Clapping)

now you can find the others
June 17th 2008, 06:16 PM
Darkhrse99

Thanks, I should have cought that;) So the answer is now right, right?
June 17th 2008, 06:28 PM
Jhevon

Quote:

Originally Posted by Darkhrse99

Thanks, I should have cought that;) So the answer is now right, right?

yeah