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

• B
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.

mfb
Mentor
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.

fresh_42
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2021 Award
Do you know how to transform degrees in radiant and vice versa?

Do you know how to transform degrees in radiant and vice versa?
Yeah.

fresh_42
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2021 Award
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.

Homework Helper
Gold Member
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.

physics user1 and Phys12
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! :)

Homework Helper
Gold Member
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.

Last edited:
Svein
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 $\sin(x)=\sin(\frac{x\cdot \pi}{180}), x>0$. One set of solutions is, of course $x=\frac{x\cdot\pi}{180}+2n\pi$, the least solution given by n=1: $x=\frac{360\cdot\pi}{180-\pi}$.

Another set of solutions is given by $x=\pi-\frac{x\cdot\pi}{180}+2n\pi$. Here the smallest solution is given by n=0: $x=\pi-\frac{x\cdot\pi}{180}$, or $x=\frac{180\cdot\pi}{180+\pi}$. This solution is the smallest value for x.

mfb
Mentor
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?