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Math Help - Where do I start this triq problem?

  1. #1
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    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} ?



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    Jason
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    Quote Originally Posted by Darkhrse99 View Post
    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
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    Sorry about that. I'll make sure to write it as Tan (x) as that is the variable we are using.
    Last edited by Darkhrse99; June 17th 2008 at 05:16 PM.
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  4. #4
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    Does Cos(x) = \frac{\sqrt17} {4} and  tan (x) =\frac{1} {\sqrt17}?
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    Quote Originally Posted by Darkhrse99 View Post
    Does Cos(x) = \frac{\sqrt17} {4} and  tan (x) =\frac{1} {\sqrt17}?
    Nope. where did 17 come from?
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  6. #6
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    Sin (x)was  \frac{1} {4} so  1^2+ 4^2=17,          x={\sqrt17}? Adjacent = {\sqrt17}
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    Quote Originally Posted by Darkhrse99 View Post
    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
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    so it is cos (x)=\frac{\sqrt15} {4} and tan (x)= \frac{1}{\sqrt15}?
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    Quote Originally Posted by Darkhrse99 View Post
    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
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  10. #10
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    CSC (t) is in the 4th quad, so cos(x)=\frac{\sqrt15} {4} and Tan (x)=\frac{-1} {\sqrt15}
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    Quote Originally Posted by Darkhrse99 View Post
    CSC (t) is in the 4th quad, so cos(x)=\frac{\sqrt15} {4} and Tan (x)=\frac{-1} {\sqrt15}
    correct

    now you can find the others
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  12. #12
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    Thanks, I should have cought that So the answer is now right, right?
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    Quote Originally Posted by Darkhrse99 View Post
    Thanks, I should have cought that So the answer is now right, right?
    yeah
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