# Where do I start this triq problem?

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• Jun 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
• Jun 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
• Jun 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.
• Jun 17th 2008, 05:12 PM
Darkhrse99
Does $Cos(x) = \frac{\sqrt17} {4}$ and $tan (x) =\frac{1} {\sqrt17}$?
• Jun 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?
• Jun 17th 2008, 05:48 PM
Darkhrse99
Sin (x)was $\frac{1} {4}$ so $1^2+ 4^2=17, x={\sqrt17}$? Adjacent = ${\sqrt17}$
• Jun 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: $\sin^2 x + \cos^2 x = 1$
• Jun 17th 2008, 05:55 PM
Darkhrse99
so it is $cos (x)=\frac{\sqrt15} {4}$and $tan (x)= \frac{1}{\sqrt15}$?
• Jun 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
• Jun 17th 2008, 06:11 PM
Darkhrse99
CSC (t) is in the 4th quad, so $cos(x)=\frac{\sqrt15} {4} and Tan (x)=\frac{-1} {\sqrt15}$
• Jun 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
• Jun 17th 2008, 06:16 PM
Darkhrse99
Thanks, I should have cought that;) So the answer is now right, right?
• Jun 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