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B Smallest number whose sin(x) in radian and degrees is equal?

  1. Jul 2, 2016 #1
    Question: what is the smallest positive real number x with the property that the sine of x degrees is equal to the sine of x radian?

    My try: 0.

    But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides of sin(theta)=sin(x), but that didn't help.
     
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  3. Jul 2, 2016 #2

    mfb

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    It works, but first you have to figure out where approximately your solution is (did you make a sketch?), and express theta in terms of x or vice versa.
     
  4. Jul 2, 2016 #3

    fresh_42

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    Do you know how to transform degrees in radiant and vice versa?
     
  5. Jul 2, 2016 #4
    Yeah.
     
  6. Jul 2, 2016 #5

    fresh_42

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    Then write it down. Put a sin before both, recognize that the sin could be canceled by a arcsin and your answer should pop up.
     
  7. Jul 2, 2016 #6

    Charles Link

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    What you are asking for is a non-zero solution ## \sin(x)=\sin(x \pi/180) ## where the ## \sin ## function in both cases now acts on radians. Suggest you graph y vs. x for both and see where they intersect.
     
  8. Jul 2, 2016 #7
    Got it, thanks! :)
     
  9. Jul 2, 2016 #8

    Charles Link

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    Because the first function is cycling at a rate of almost 60 times the second function, the first nonzero intersection is going to occur near (just before) ## x=\pi ##. The reason is the second function is going up very slowly and the first one has made it back down to zero already. If you let ## x=\pi-\Delta ##, you can use trig identities and Taylor series. (## \theta=(\pi-\Delta)\pi/180 ## is small so that e.g. ## \sin(\theta)=\theta ## approximately.) This gives ## \sin(\pi-\Delta)= \sin(\Delta)=\Delta=(\pi- \Delta )\pi /180 ## which gives ## \Delta=\pi^2/180 ## (approximately) so that ## x=\pi- \pi^2/180 ## is a good approximate solution for ## x ##. Further refinement could be done with higher order terms or by plugging into a calculator near this point.
     
    Last edited: Jul 2, 2016
  10. Jul 3, 2016 #9

    Svein

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    To spell it out: You seek the values of x, where [itex] \sin(x)=\sin(\frac{x\cdot \pi}{180}), x>0[/itex]. One set of solutions is, of course [itex]x=\frac{x\cdot\pi}{180}+2n\pi [/itex], the least solution given by n=1: [itex] x=\frac{360\cdot\pi}{180-\pi}[/itex].

    Another set of solutions is given by [itex] x=\pi-\frac{x\cdot\pi}{180}+2n\pi[/itex]. Here the smallest solution is given by n=0: [itex]x=\pi-\frac{x\cdot\pi}{180} [/itex], or [itex]x=\frac{180\cdot\pi}{180+\pi} [/itex]. This solution is the smallest value for x.
     
  11. Jul 3, 2016 #10

    mfb

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    Why approximating better than "a bit below pi", if the analytic solution can be found by solving a linear equation?
     
  12. Jul 3, 2016 #11

    Charles Link

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    Svein's solution(s) are rather clever=I totally overlooked them. In reading them, I wasn't surprised that his first one doesn't pick up the smallest value for x, but rather picks up the intersection near ## x=2 \pi ##. His second one gets the exact answer by solving a linear equation, but without knowing that the first one did not get the answer near ## x=\pi ##, that set of solutions could easily be overlooked. I did not expect an exact answer to this one. Very good solution by @Svein. ...editing... In post #8, had I kept the next term when I solved for ## \Delta ##, I would have had the exact solution (given by Svein) without even knowing that it was the exact answer...
     
    Last edited: Jul 3, 2016
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