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

[Phys12]

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]

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]

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

 

[Phys12]

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

Yeah.

 

[fresh_42]

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.

 

[Charles Link]

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.

 

[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! :)

 

[Charles Link]

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.

 

[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]

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?

 

[Charles Link]

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...