Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I see no trig function for this

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data

    244.10746 = 845.9064sin(θ) - 274.87cos(θ)

    2. Relevant equations

    Is there? I have no clue.

    3. The attempt at a solution

    I plugged it into WolframAlpha and I got an incomprehensible answer.
  2. jcsd
  3. Jun 7, 2012 #2
    Have you attempted to use the property [itex]1 = sin^{2}(\theta) + cos^{2}(\theta)[/itex] to get an equation in terms of only sin or cos?
  4. Jun 7, 2012 #3
    But I can't use that trig function for this equation.
  5. Jun 7, 2012 #4
    Try using sin(A-B) = sinA cosB - cos A sinB. Start out by dividing both sides of the equation by sqrt (845.90642 + 274.872)

  6. Jun 7, 2012 #5
    Of course you can. Let [itex]x = \cos \theta[/itex]. Then [itex]\sin \theta = \sqrt{1-x^2}[/itex]. All it takes is a little more algebra to massage this into a polynomial equation in [itex]x[/itex] that you can then convert back into [itex]\theta[/itex].
  7. Jun 7, 2012 #6
    I got x = 809.919 after plugging everything in. But I can't use that x-value for the sin(θ) = √(1-x^2) because that would become a negative number inside of that root.
  8. Jun 7, 2012 #7
    I suspect you did some algebra wrongly; I get a value for x that is between -1 and 1 as required. Why don't you type out a couple steps of algebra?
  9. Jun 8, 2012 #8
    I must be squaring it wrong. This time I got x = 0.91059, but I know it's supposed to be 0.82917.

    244.10746 = 845.9064sin(θ) - 274.87cos(θ)

    244.10746 = 845.9064[itex]\sqrt{1-x^2}[/itex] - 274.87x

    *multiply everything by the power of 2*

    59588.4520 = 715557.6376 - 715557.6376x2 - 75553.5169x2

    I'm guessing I squared the above equation incorrectly...
  10. Jun 8, 2012 #9
    Yes, you did. On the right hand side, you've squared each term, but that's not how squaring works.

    [tex](a+b)^2 = a^2 + 2ab + b^2 \neq a^2 + b^2[/tex]
  11. Jun 8, 2012 #10
    But then if I square it, I would still be left with [itex]\sqrt{1-x^2}[/itex], wouldn't I?
  12. Jun 8, 2012 #11
    Yes. How might you isolate that square root so you don't get any cross terms when you square both sides?
  13. Jun 8, 2012 #12
    I put the square root by itself and got this:

    [itex]\sqrt{1-x^2}[/itex] = [itex]\frac{244.10746 + 274.87x}{845.9064}[/itex]

    I did the calculations wrong again, and got this:

    0 = 70.4433 + 158.6412x + 90.3166x2

    It's probably because I didn't square the 845.9064 as well.
  14. Jun 8, 2012 #13
    Well, once you're able to square these numbers correctly with a calculator, you should be getting close to something that you an arrange and solve by the quadratic formula.
  15. Jun 9, 2012 #14
    I'm surprised that you guys didn't even try to solve this problem using the much simpler method that I alluded to in Response #4 (which does not even require solving a quadratic equation). Here is an additional hint:

    A sinθ - B cosθ = sqrt(A2 + B2) (sinθ cos[itex]\phi[/itex] -cos[itex]\theta[/itex] sin[itex]\phi[/itex]) = sqrt(A2 + B2) sin([itex]\theta[/itex]-[itex]\phi[/itex])

    where [itex]\phi[/itex] = arctan (B/A)

  16. Jun 9, 2012 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Here's where Chestermiller's suggestion comes from. The idea is to find R and φ such that
    A &= R \cos \varphi \\
    B &= R \sin \varphi
    \end{align*} so that
    A \sin\theta - B \cos \theta &= (R \cos\varphi)\sin\theta - (R\sin\varphi)\cos\theta \\
    &= R(\cos\varphi\sin\theta - \sin\varphi\cos\theta) \\
    &= R \sin(\theta-\varphi).
    \end{align*} If you divide the second equation by the first, you get
    $$\frac{B}{A} = \frac{R\sin\varphi}{R\cos\varphi} = \tan\varphi,$$ and using the Pythagorean identity, you get
    $$1 = \cos^2 \varphi + \sin^2\varphi = \left(\frac{A}{R}\right)^2 + \left(\frac{B}{R}\right)^2,$$ or equivalently, ##R^2 = A^2+B^2##.

    This trick is a good one to know. As he noted, it's a lot less tedious to solve the problem using this method than dealing with the quadratic equation, especially when you have non-integer numbers like the ones in this problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook