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One More Trig Identity Problem

  1. Dec 2, 2006 #1
    There is one last problem i have on my trig assignment and i have no clue how to do it. The questions is:

    Find sinx, cosx, tanx, sin2x, cos2x, and tan2x from the given information:

    secx=5, sin is negative.

    If anyone could show me how to do this problem it would be soooo appreciated.

  2. jcsd
  3. Dec 2, 2006 #2
    [tex] \sec x = 5 \Rightarrow \frac{1}{\sec x} = \frac{1}{5} = \cos x [/tex]

    [tex] \cos x [/tex] is positive and [tex] \sin x [/tex] is negative in the fourth quadrant.

    [tex] \sin 2x = 2\sin x \cos x [/tex]

    [tex] \cos 2x = 1-2\sin^{2} x [/tex]
    Last edited: Dec 2, 2006
  4. Dec 2, 2006 #3
    so how do i find sinx cause if you go from cos wouldnt you have to express sin in terms of tan since cos=sin/tan?
  5. Dec 2, 2006 #4
    draw a right triangle. If you know the adjacent side is 1, and the hypotenuse is 5, then the opposite side is [tex] 2\sqrt{6} [/tex].
    Last edited: Dec 2, 2006
  6. Dec 2, 2006 #5
    I dont understand how that will help?
  7. Dec 2, 2006 #6
    not that im saying your wrong lol im just saying i dont understand how to do that
  8. Dec 2, 2006 #7
    ah i get you
  9. Dec 2, 2006 #8
    wouldnt the other side be square root of 24 if you use pythagorins theorem?
  10. Dec 2, 2006 #9
    yes it would. sorry, a typo.
  11. Dec 2, 2006 #10
    Ok just to make sure im doing it correctly here are the answers i got, would be great if you could tell me if i totally messed it up lol.

    I got:

    sinx= -2 squareroot 6/5
    tanx= -2 squareroot 6
    sin2x= -4 squareroot 6/25
    cos2x= 23/25
    tan2x= -4 squareroot 6/5

    thanks for all the help!
  12. Dec 2, 2006 #11
    [tex] \cos 2x = -\frac{23}{25} [/tex]

    [tex] \tan 2x = \frac{\sin 2x}{\cos 2x} [/tex]

    All the rest look good.
  13. Dec 2, 2006 #12
    why are those 2 like that?
  14. Dec 2, 2006 #13
    how can cosx and cos2x both be 1/5?

    Also heres how i did those 2 questions im just wondering why they are wrong?

    =1-2(-2 squareroot 6/5)squared

    =2(-2 squareroot 6)/1-6
    =-4 squareroot 6/-5
    =4 squareroot 6/5
  15. Dec 2, 2006 #14
    I said that it should be a -23/25.

    And you are right.
    Last edited: Dec 2, 2006
  16. Dec 5, 2006 #15
    To quote courtigrad:
    "[tex] \sin 2x = 2\sin x \cos x [/tex]"

    I think this might be a problem for some, because I was doing problems similar to this the other day, and only today did we begin learning the double-angle, half-angle, and power down formulas.
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