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Homework Help: Simultaneous trigonometric equations

  1. Apr 10, 2008 #1
    0=-85.7cos(theta)+bcos30-85.7
    0=-85.7sin(theta)+bsin30
    Can someone please tell me what I am doing wrong.
    theta=sin^-1(bsin30/85.7)
    then sub that into the first equation.
    I then solved the equation on my calculator and it gave me 9999999999 which i know is wrong.
     
  2. jcsd
  3. Apr 11, 2008 #2
    Are you trying to solve the equation for b or theta? It's hard to tell.
     
  4. Apr 11, 2008 #3
    i am trying to solve it for b
     
  5. Apr 11, 2008 #4

    tiny-tim

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    Hi Ry122! :smile:

    Hint: eliminate θ by using cos²θ + sin²θ = 1. :smile:
     
  6. Apr 11, 2008 #5
    Move the theta terms in both equations to the left hand side.

    Square both equations.

    Add the two that you get, now you can use tiny-tim's trick.

    What you have left is a quadratic in b, solve it. I'm guessing somewhere along the way you assumed that b > 0 since every else you explicitly took care of the signs, you can use this to throw away a spurious solution if that is the case (leaving you with only one value for b).

    Now take either one of the two original equations and plug b into it, and then solve for theta.

    Ok that was a complete set of instructions instead of a hint, but this is after all just mathematical gymnastics and not a meaningful exercise.
     
  7. Apr 11, 2008 #6
    Is this right?
    (85.7cos(theta)+85.7sin(theta)=bcos30-85.7+bsin30)^2
    7344.49cos^2(theta)+7344.49sin^2(theta)=b^2cos^2(90)-(85.7)^2+b^2sin90
    1=(b^2cos^2(90)-(85.7)^2+b^2sin90)/7344.49x7344.49
     
    Last edited: Apr 11, 2008
  8. Apr 11, 2008 #7
    i cant solve it because theres still a cos^2 and a sin^2. how do i get rid of them?
     
  9. Apr 11, 2008 #8

    tiny-tim

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    No no no … you added, and then squared. :frown:

    (btw, why didn't you just copy my θs and ²s … it makes reading much easier?)

    Square first, then add!

    Try again! :smile:
     
  10. Apr 11, 2008 #9
    (85.7)²cos²θ=b²cos²90-(85.7)²
    (85.7)²sin²θ=b²sin²90

    (85.7)²cos²θ+(85.7)²sin²θ=b²cos²90-(85.7)²+b²sin²90
    this is the same isnt it? or am i adding incorrectly?
     
    Last edited: Apr 11, 2008
  11. Apr 11, 2008 #10

    tiny-tim

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    No no no … you've put (bcos30 - 85.7)² = (bcos30)² - (85.7)². :cry:

    What should it be?

    And simplify the left-hand side … that was the whole point … ! :smile:

    (oh, and (bcos30)² is b²cos²30. not b²cos²90 :rolleyes:)
     
  12. Apr 11, 2008 #11
    should it be
    bcos30=b.866
    (b)²(.866)²

    should - (85.7)² be (-85.7)^2?
     
    Last edited: Apr 11, 2008
  13. Apr 11, 2008 #12
    sorry i understand what you are saying now
     
  14. Apr 12, 2008 #13

    tiny-tim

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    Hi Ry122! :smile:

    Are you ok now on this?

    If so, hit "[SOLVED]" under "Thread Tools" on the menu bar. :smile:
     
  15. Apr 12, 2008 #14
    I still need some help.
    (85.7cos(theta))²=(bcos30-85.7)² eq1
    (85.7sin(theta))²=(bsin30)² eq2
    85.7²cos²(theta)=b²cos²30-85.7bcos30-85.7bcos30+85.7² eq1
    simplified
    85.7²cos²(theta)=b²cos²30-171.4bcos30+85.7²
    eq2
    (85.7sin(theta))²=(bsin30)²
    85.7²sin²(theta)=b²sin²30

    add them
    85.7²sin²(theta)+85.7²cos²(theta)=b²sin²30+b²cos²30-171.4bcos30+85.7²
    sin²(theta)+cos²(theta)=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
    1=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
    Is this correct?
    I dont know how to solve it with that sin squared and cos squared still in there.
     
    Last edited: Apr 12, 2008
  16. Apr 12, 2008 #15
    ok i solved it. I got 39.152. Can someone please tell me if this is the correct answer? It would be greatly appreciated.
     
  17. Apr 12, 2008 #16
    If you have values for both b and theta, then just check it yourself-- plug into both equations, if both equations are valid then you found the correct solution.
     
  18. Apr 12, 2008 #17

    tiny-tim

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    Hi Ry122! :smile:

    Sorry … no!

    First, where did your "^4" come from??

    Second, the whole point of adding equal amounts of cos²θ and sin²θ (btw, why did you stop copy-and-pasting the ²θ?) was so that you could replace cos²θ + sin²θ by 1 in the very next line!

    Then you get 85.7² on both sides, so they cancel out (in this case … they wouldn't normally).

    Try again! :smile:
     
  19. Apr 12, 2008 #18
    i had to get rid of the 2 85.7² on the left side so i moved them to the denominator on the right side


    85.7²sin²(theta)+85.7²cos²(theta)=b²sin²30+b²cos²3 0-171.4bcos30+85.7²


    85.7² x 85.7² = 85.7^4

    sin²(theta)+cos²(theta)=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4

    then use the identity

    1=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
     
    Last edited: Apr 12, 2008
  20. Apr 12, 2008 #19

    tiny-tim

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    Nooo … it should only be /(85.7)² at the end! :rolleyes:

    Now try it! :smile:
     
  21. Apr 12, 2008 #20
    can you tell me why?
     
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