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

In summary, the smallest positive real number x with the property that the sine of x degrees is equal to the sine of x radian is approximately x = 3.1416 - 0.0174533 = 3.1241467, or more precisely, x = 180 * pi / (180 + pi) = 180 / (1 + pi / 180). This can also be solved analytically as x = pi / 180.
  • #1
Phys12
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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
Phys12 said:
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
Do you know how to transform degrees in radiant and vice versa?
 
  • #4
fresh_42 said:
Do you know how to transform degrees in radiant and vice versa?
Yeah.
 
  • #5
Phys12 said:
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
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
Charles Link said:
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
Phys12 said:
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
Charles Link said:
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
Charles Link said:
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
mfb said:
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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1. What is the smallest number whose sin(x) in radians and degrees is equal?

The smallest number whose sin(x) in radians and degrees is equal is 0. This is because sine is a periodic function with a period of 2π radians, or 360 degrees. Therefore, any multiple of 2π or 360 will have a sine value of 0.

2. How do you calculate the smallest number whose sin(x) in radians and degrees is equal?

To calculate the smallest number whose sin(x) in radians and degrees is equal, you can use the inverse sine function (arcsine) on a calculator. Simply enter 0 as the input and solve for x. The resulting value will be the smallest number with an equal sine value in radians and degrees.

3. Is there a mathematical formula to find the smallest number whose sin(x) in radians and degrees is equal?

Yes, there is a mathematical formula to find the smallest number whose sin(x) in radians and degrees is equal. It is x = 2πn, where n is any integer. This formula represents all the possible angles that have a sine value of 0, which is the smallest possible value.

4. Can the smallest number whose sin(x) in radians and degrees be negative?

No, the smallest number whose sin(x) in radians and degrees is equal must be positive. This is because sine is a positive value for angles between 0 and 90 degrees, and 0 and π/2 radians. Any negative angle would result in a negative sine value, which is not equal to 0.

5. Why is it important to understand the concept of the smallest number whose sin(x) in radians and degrees is equal?

Understanding this concept is important in trigonometry and other fields that use sine and cosine functions. It helps in solving equations and finding the solutions for right triangles and other geometric problems. It also provides a better understanding of the periodic nature of trigonometric functions.

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