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Homework Help: Simultaneous equation involving cos, sin

  1. Jun 23, 2016 #1
    1. The problem statement, all variables and given/known data

    i have solve a engineering problem until this part i'm stuck

    C cos 30 - 1.375 cosθ = 0 ----(1)
    C sin 30 + 1.375 sin θ = 15 ---(2)

    . i have 2 unknown C and θ, however the change to cos to sin or via versa make me lost because of my poor mathematic

    2. Relevant equations

    3. The attempt at a solution

    i have try changing the sin θ to √(1-cos 2θ)/2 but i still did not get the right answer
    i hope that i can have a guide to solve this by using equation funtion in scientific calculator or do it manually.

    TQ in advance
  2. jcsd
  3. Jun 23, 2016 #2

    Buzz Bloom

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    Hi urazi:

    Use (1) to get an expression for C. Then substitute this for C in (2).
    You now have an equation of the form:
    A cos θ = B sin θ + 15.​
    Now square both sides and use
    cos2 θ = 1 - sin2 θ.​
    You then have a quadratic equation involving the variable sin θ.

    Hope this helps.

  4. Jun 23, 2016 #3

    Ray Vickson

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    Are you sure that your equations are correctly written? Your equations do not have any real solutions. You can verify this fact for yourself: just use your first equation to solve for ##C## in terms of ##\cos \theta##, then substitute that expression into the second equation, to get an equation involving ##\sin \theta## and ##\cos \theta## alone (no ##C##). Plot the left-hand-side of that equation for ##0 \leq \theta \leq 360^o## and you will see that there are no real values of ##\theta## that can possibly work.

    You could do more: as suggested in #2, substitute ##\sin \theta = \sqrt{1 - \cos^2 \theta}## or ##\sin \theta = - \sqrt{1 -\cos^2 \theta}## to obtain (two different) quadratic equations in ##x = \cos \theta##. You will find that they do not have real solutions.
  5. Jun 23, 2016 #4


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    Hello urazi. Welcome to PF !

    As Ray pointed out, there is no real solution for this system of equations.

    I looked into this a bit further. It looks to me that in order for there to be real solutions, the constant on the right hand side of Eq. 2, must be between approximately ±2.52 .

    Perhaps there's a typo and that should be 1.5 rather than 15 .

    Beyond that:
    Another way to solve this system is to:
    1. Solve Eq. 1 for 1.375 cosθ , then square both sides.
    2. Solve Eq. 2 for 1.375 sinθ , then square both sides.
    3. Add the two resulting equations, noting that sin2θ + cos2θ = 1
    This will result in a an equation that's quadratic in C .
  6. Jun 23, 2016 #5
    thank you buzz,
    yes, i have check again that 15 is should be 1.5 if not i will not get = 0 for the second equation
    i have try and i get theta = 56 and C = 0.88 using your solution,
    before this i have try and get theta 46 and C=0.69
    however the ans i should get is theta = 40.9 and C =1.2
  7. Jun 23, 2016 #6
    thank you, yes the equation should be 1.5 not 15 .
  8. Jun 23, 2016 #7

    yes the equation should use 1.5 rather than 15
  9. Jun 23, 2016 #8


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    So, have you been able to solve it using any of the suggestions given ?
  10. Jun 23, 2016 #9

    Hai sammy, i have try using your way
    i got theta =46 and C =1.1
    however the answer i should get is theta = 40.9 and C is 1.2

  11. Jun 23, 2016 #10
    opss no i 'm using your way i get C^2 = -0.718
  12. Jun 23, 2016 #11


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    I get C = 1.2 . That's approximate. Actually I get C = 1.2006939...

    That's one of two answers given by the quadratic formula.
  13. Jun 23, 2016 #12


    am i doing it right?...sorry for troubling you

    1.891cos^2= - 0.75 c^2 ----(1)
    1.891sin^2= - 0.25 c^2 +2.25------ (2)

    adding 1 to 2

    1.891 (1) = - 0.75 c^2 - 0.25 c^2 +2.25
    C^2-0.359 =0
  14. Jun 23, 2016 #13


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    Well, you did comment earlier that your math skills aren't too good .

    (1) Can't have a negative sign. It's the result of squaring .

    (2) What do you get if you square ( 1.5 - C/2) ?
  15. Jun 24, 2016 #14

    Ray Vickson

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    You are doing it the hard way. Much easier: solve for C in terms of cos(θ) from the first equation, then substitute that into the second equation to get an equation of the form a*cos(θ) + b*sin(θ) = R. Here, R = 1.5, b = 1.375 and a = 1.375*tan(30°). Now just re-express a * cos(θ) + b* sin(θ) in the form A*cos(θ+k), a single trig term with amplitude A and phase k, both of which are easily computed. [If you prefer, you could write, instead, A*sin(θ+r), with the same amplitude A but a different phase r.]
    There is a standard trick for finding A and k or r, and it is used very, very widely in engineering applications.

    Anyway, it is now easy to deal with the equation A*cos(θ+k) = R or A*sin(θ+r) = R.
  16. Jun 25, 2016 #15
    thank you very much to sammy and ray. and all who respond....i have solved the problem :) :)
  17. Jun 25, 2016 #16
    and thank you very much for your patience to guide me.......
  18. Jun 25, 2016 #17


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    @urazi ,

    Whether some particular approach is relatively easier or more difficult depends upon many things.

    As Ray said, any expression of the form, a⋅cos(θ) + b⋅sin(θ), can be expressed as purely a sine or as purely a cosine, either one with some "phase shift". I'm not sure if the following is the "trick" Ray was thinking of, but I find it's a good way to break this down.
    Start with the expression: a⋅cos(θ) + b⋅sin(θ) .
    Define A as ##\ A=\sqrt{a^2+b^2\,}\ .\,## Then A is the length of the hypotenuse of a right triangle with legs of length |a| and |b| .
    Write our expression as ##\ A\left(\frac aA⋅\cos(θ) + \frac bA⋅\sin(θ)\right) \ .##
    Then define angle ##\phi## so that either ##\ \sin(\phi)=\pm\frac aA \text{ and } \cos(\phi)=\frac bA\ ##
    or ##\ \cos(\phi)=\frac aA \text{ and } \sin(\phi)=\pm\frac bA\ ##
    (Choose sings as convenient.)
    The first choice for ##\phi## gives the angle addition formula for sine, the second choice for ##\phi## gives the angle addition formula for cosine.

    In the first case we get ##\ \sin(\phi)⋅\cos(θ) + \cos(\phi)⋅\sin(θ) = \sin(θ+\phi)\ .## The other case works similarly for cosine.​

    While the above can very useful on many occasions, in my opinion, in the case of the problem given here, I think the method I outlined in post #4 works out very simply.

    By The Way: You can solve this system with a minimum of algebra by using a graphing calculator. Solve each equation for C. Each gives C as a function of θ. Graph both of these (C becomes Y1 and Y2, θ becomes X) and use the calculator to solve for the two points of intersection.
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