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Trig help

  1. Oct 31, 2005 #1
    Ive been given very little notes from my teahcer, and cant do these. looked online, but its was no use.

    1. Using the identities to find the vales of each expression (no calc.)
    (i) sin t = 15/17 , cos t = 8/17 find remaining trig functions
    (ii) sec^2 Pie/12 - tan^2 Pie/12

    2. Find exact value
    (i) tan* = -3 , cos *>0

    3.Use reference angles to find exact value (no calc.)
    (i) cos 225*
    (ii) sin ( -pie/6 )


    Thanks a bunch!
     
  2. jcsd
  3. Oct 31, 2005 #2

    VietDao29

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    Homework Helper

    >_<, Ahhh, you didn't show us any of your work or even your thought (how do you think you can approach the problem, and where you get stuck!).
    You can click here to find some more information about 'old' posting rules or you can see a new version of it here.
    -------------------
    Anyway, I'll give you a couple of hints.
    For #1, what are the trig functions? How are they defined? Something like:
    [tex]\tan x = \frac{\sin x}{\cos x}[/tex],...
    You already have sint t, cos t, can you find out the others?
    ---
    [tex]\sec ^ 2 \frac{\pi}{12} - \tan ^ 2 \frac{\pi}{12}[/tex]. You should know that:
    [tex]\sec x = \frac{1}{\cos x}[/tex], now just convert everything to sin and cos function, then simplify the expression and you will get something really neat.
    Note that:
    sin2x + cos2x = 1 or 1 - sin2x = cos2x.
    ------------
    For #2,
    You know that tan x = -3 < 0, cos x > 0, now you need something that relates the tan function and the cos function, what equation is that?
    Isn't it:
    [tex]1 + \tan ^ 2 x = \frac{1}{\cos ^ 2 x}[/tex]?
    (Do you know how to arrive at that equation?)
    Note that cos x > 0.
    -----------------
    For #3,
    What's [tex]\sin \frac{\pi}{6}[/tex]? So can you find out what [tex]\sin -\frac{\pi}{6}[/tex] is?
    You can do the same for cos 255o.
    Viet Dao,
     
  4. Oct 31, 2005 #3
    so for question 1(i) the answer is....tan t = 1.87* ?

    1(ii)
    1/cos Pie/12 - sinx/cosx pie /12

    1 - sin2x = cos2x.

    so...

    1 - sin x / cosx (pie/12)^2 = (1/cosx (pie/12))^2

    ahhh...im so lost

    #2...dont have a sweet clue...i have no notes on this...kinda clueless

    #3 is it
    cos225 x 180/pie?

    pie225/180pie?

    =1.25?
     
  5. Oct 31, 2005 #4

    VietDao29

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    Homework Helper

    (i) There are other trig functions as well, what are they?
    sine, cosine, tangent, cotangent, secant, cosecant.
    Can you find them?
    By the way, I'd write it as tan t = 15 / 8.
    Note that tan t = 15 / 8 not tan t = 15 / 8o.
    (ii) The bolded part is wrong.
    You should note that:
    [tex]\cos ^ 2 x \neq \cos (x ^ 2)[/tex]
    I'll help you a bit:
    [tex]\sec ^ 2 x - \tan ^ 2 x = \frac{1}{\cos ^ 2 x} - \frac{\sin ^ 2 x}{\cos ^ 2 x} = \frac{1 - \sin ^ 2 x}{\cos ^ 2 x} = \frac{\cos ^ 2 x}{\cos ^ 2 x} = ...[/tex].
    So what's:
    [tex]\sec ^ 2 \frac{\pi}{12} - \tan ^ 2 \frac{\pi}{12}[/tex]?
    Here's a quick proof:
    [tex]1 + \tan ^ 2 x = 1 + \frac{\sin ^ 2 x}{\cos ^ 2 x} = \frac{\sin ^ 2 x + \cos ^ 2 x}{\cos ^ 2 x} = \frac{1}{\cos ^ 2 x}[/tex].
    Now, you have tan x = -3.
    Rearrange the equation a bit, you will have:
    [tex]\cos ^ 2 x = \frac{1}{1 + \tan ^ 2 x}[/tex]. Can you go from here?
    Note that cos x > 0.
    Nope, the cos function, and sin function will always return value in [-1, 1], and [itex]1.25 \notin [-1,\ 1][/itex].
    225o is a measure unit for angle.
    You can do it as:
    cos(225o) = cos(180o + 45o) = ...
    Can you go from here?
    --------------------------
    P.S: It's high recommended that you go back and re-read the book. :wink:
    Viet Dao,
     
    Last edited: Oct 31, 2005
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