1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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