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Math Help - Exact value for trig ratios

  1. #1
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    Exact value for trig ratios

    I have a question I'm having a bit of trouble with lads,

    Determine the exact values for the 6 trig ratios Θ(theta) lies in the standard position with its terminal arm passing through the point P(-3,-5).

    I know I'm supposed to find Sin, Cos, Tan, Csc, Sec, and Cot, but how do I find the radian measure so I can do so?

    If anyone can help it would really be great, thanks so much!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Random-Hero- View Post
    I have a question I'm having a bit of trouble with lads,

    Determine the exact values for the 6 trig ratios Θ(theta) lies in the standard position with its terminal arm passing through the point P(-3,-5).

    I know I'm supposed to find Sin, Cos, Tan, Csc, Sec, and Cot, but how do I find the radian measure so I can do so?

    If anyone can help it would really be great, thanks so much!
    to begin, did you actually draw the diagram as described?
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  3. #3
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    Of course,



    I'm assuming I have to find the angle in blue in radians, correct? Now how do I go about finding that value?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Random-Hero- View Post
    Of course,



    I'm assuming I have to find the angle in blue in radians, correct? Now how do I go about finding that value?
    well, this diagram isn't complete. you need to draw a vertical line from the point where the red line touches the circle up to the x-axis. do you see the right triangle? now, do you remember SOHCAHTOA? (sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent)

    do you also recall that sec(x) = 1/cos(x), csc(x) = 1/sin(x) and cot(x) = 1/tan(x) ?

    do you also recall (haha, getting sick of that phrase, i bet) that in the third quadrant, only tangent and cotangent are positive, while all the other trig ratios are negative?
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  5. #5
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    I managed to find that angle (at least I think I found it properly). First I foun the hyp. So I did (-5)^2 + (-3)^2 rooted gave me root34, then I did Sin=Opp/Hyp which was sin=(-5)/root34 which ultimately gave me -59 degrees. So I then converted that to radians, and couldn't really reduce, so know I'm stuck with 59pi/180 (or is it -59pi/180?)

    So now do I find the remaining angle on the other side of the red line? and use that to find my 6 values?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Random-Hero- View Post


    I managed to find that angle (at least I think I found it properly). First I foun the hyp. So I did (-5)^2 + (-3)^2 rooted gave me root34, then I did Sin=Opp/Hyp which was sin=(-5)/root34 which ultimately gave me -59 degrees. So I then converted that to radians, and couldn't really reduce, so know I'm stuck with 59pi/180 (or is it -59pi/180?)

    So now do I find the remaining angle on the other side of the red line? and use that to find my 6 values?
    if you listened to what i said, you would realize that finding the angle is completely unnecessary. you found that the hypotenuse is \sqrt{34}, this is correct.

    now, \sin \theta = - \frac {5}{\sqrt{34}}

    \cos \theta = - \frac {3}{\sqrt{34}}

    \tan \theta = \frac 53

    just by following the formulas i gave you and the directions on what to make negative and positive. see my first post. as you see, you don't need to know what \theta is. now finish up
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