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Trig questions

  1. Nov 8, 2005 #1

    dnt

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    im trying to remember how to do some trig.

    first of all, how do you find cos^-1 (5/3) without using a calculator?

    2nd, i need help proving this identity:

    2sin^2(x) - cos (x) - 2 sin (x) + 1 = 0

    3rd, how do you evaluate sin 165 (which is sin (11/12 pi)) again without using a calculator.

    thanks for the help. i know im suppose to try them first but i havent done this in a while and im really stuck. :confused:

    also if someone could give me a good trig link so i can refresh on my trig, id appreciate it.
     
  2. jcsd
  3. Nov 8, 2005 #2

    benorin

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    Homework Helper

    sin 165 = sin(120+45), then use formula for sin(A+B)

    recall that sin^2 x + cos^2 x = 1

    and cos^-1 (5/3) does not exist since -1 <= cos(x) <= 1 for any x
     
  4. Nov 8, 2005 #3

    HallsofIvy

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    As benorin said, cos(5/3) is impossible. cosine of any number cannot be larger than 1.
    (2) is also impossible: it is NOT an identity. In particular, for x= 45 degrees,
    sin 45= cos 45= [itex]\frac{\sqrt{2}}{2}[/itex] so
    2 sin^2(45)- cos(45)- 2sin(45)+ 1=
    [tex]\frac{2}{2}- \frac{\sqrt{2}}{2}- \sqrt{2}+ 1= 2- 3\frac{\sqrt{2}}{2}[/tex]
    not 0.
    For (3), sin(165)= sin(120+ 45)= sin(2(60)+ 45). You should know the sine and cosine of both 60 and 45 and use the sum formula and double angle formula.
     
  5. Nov 8, 2005 #4

    dnt

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    ok, so it would be:

    (sin 120)(cos 45) + (cos 120)(sin 45)

    correct?


    the full problem was sin[2 cos^-1(5/3)]

    so that has no answer?
     
  6. Nov 8, 2005 #5

    dnt

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    i know how to do it if you use sin (120 + 45), assuming my work in the post right above is correct, but just for learning sake, how do you do it using sin (2(60) + 45) - the double angle formula with the sum formula?

    edit: would you say [sin (2(60))* cos(45)] + [cos (2(60))* sin (45)] and then use the double angle formula to get:

    (2)(sin 60)(cos 60) * (cos 45) + (cos 60)^2 - (sin 60)^2 * (sin 45)
     
    Last edited: Nov 8, 2005
  7. Nov 8, 2005 #6

    dnt

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    another identity i could use some help in:

    (csc X)/(1 - cos X) = (1 - sin X)/(cos^2 X)

    what i tried to do was say cos^2 X = 1 - sin^2 X

    then i factored it so it would cancel with the top. leaving me with:

    (csc X)/(1 - cos X) = 1/(1 + sin X)

    but now im stuck.
     
  8. Nov 8, 2005 #7

    HallsofIvy

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    The problem you have is that that is NOT an identity. Again, if X= 45 degrees, then the formula becomes
    [tex]\frac{\sqrt{2}}{1- \frac{\sqrt{2}}{2}}= \frac{1- \sqrt{2}}{\frac{1}{2}}[/tex]
    That is NOT true! Where are you getting these problems?
     
  9. Nov 8, 2005 #8

    dnt

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    a book. well i feel better for not being able to get them :)

    thanks for the help.
     
  10. Nov 9, 2005 #9

    dnt

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    oops...the actual problem was

    sin[2 cos^-1(3/5)]

    (not 5/3)

    now can someone help me with this? thanks.
     
  11. Nov 9, 2005 #10
    [​IMG] I hope this helps you,
    Regards,
    Sam
     
  12. Nov 10, 2005 #11

    dnt

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    i still dont know how to find the angle (in degrees or radians) without using a calculator. please help.
     
  13. Nov 10, 2005 #12
    In this one "first of all, how do you find cos^-1 (5/3) without using a calculator? "

    It's extremely simple

    no answer!

    the ratio between the adjacent and hypotenus can never be greater than one, because the adjacent side cannot be longer than the hypotenus.
     
  14. Nov 10, 2005 #13

    dnt

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    thanks, but that was already explained above. how do you do cos^-1 (3/5)?
     
  15. Nov 11, 2005 #14
    Is the question actually asking you to find the value of arccos(3/5)?

    Or are you trying to work it out as a step in your calculations?

    Regards,
    Sam
     
  16. Nov 11, 2005 #15

    HallsofIvy

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    You said before:
    "oops...the actual problem was

    sin[2 cos^-1(3/5)]"

    BerryBoy then showed you how to get that, without actually finding the angle. In fact there is no way to get an exact answer for the cos-1(3/5).
     
  17. Nov 11, 2005 #16

    dnt

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    sorry, the full problem is sin[2 cos^-1(3/5)]

    but is there any way to find cos^-1(3/5) by itself without a calculator?

    if not, how do you find the answer to the above full problem? would it be:

    2(4/5)(3/5) = 24/25

    is that right?
     
  18. Nov 11, 2005 #17

    HallsofIvy

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    There is no way, other than by approximation, either by Taylor's series or by successive approximations.

    Yes,you can do sin(2arccos(3\5)) just as suggested. If cos x= 3/5, sinx= 4/5 sin(2x)= 2(4/5)(3/5)= 24/25.
     
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