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

  • #1
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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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  • #2
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I tried taking inverse on both sides of sin(theta)=sin(x), but that didn't help.
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.
 
  • #3
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Do you know how to transform degrees in radiant and vice versa?
 
  • #4
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Do you know how to transform degrees in radiant and vice versa?
Yeah.
 
  • #5
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Yeah.
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.
 
  • #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.
 
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  • #7
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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.
Got it, thanks! :)
 
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  • #8
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Got it, thanks! :)
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.
 
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  • #9
Svein
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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.
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.
 
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  • #10
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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.
Why approximating better than "a bit below pi", if the analytic solution can be found by solving a linear equation?
 
  • #11
Charles Link
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Why approximating better than "a bit below pi", if the analytic solution can be found by solving a linear equation?
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...
 
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