Solve Arcsin(1/x)=arctan(x)=>x=?

1. Jun 3, 2009

fredrick08

1. The problem statement, all variables and given/known data
can someone please tell me why arcsin(1/x)=arctan(x)=>x=-$$\sqrt{}(-2+\sqrt{}20$$)/(-1+$$\sqrt{}5$$)?????
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 3, 2009

fredrick08

Re: arcsin(1/x)=arctan(x)=>x=?

do you just derive them, then solve for x, since they eliminate the trig functions?

3. Jun 3, 2009

fredrick08

Re: arcsin(1/x)=arctan(x)=>x=?

nope... that doesnt work...

4. Jun 4, 2009

Staff: Mentor

Re: arcsin(1/x)=arctan(x)=>x=?

I'm not sure what you mean by "derive them." The equation arcsin(1/x) = arctan(x) is not an identity (an equation that's always true); it is a conditional equation, one that is true only for certain values of x.

Let $\alpha$ = arcsin(1/x), and let $\beta$ = arctan(x). From these equations, you should see that sin($\alpha$) = 1/x and tan($\beta$ = x.
Now, draw two right triangles, with the legs and hypotenuses labelled according to the last two equations above. Since $\alpha$ = $\beta$, the two triangles must be similar (but not necessarily congruent), which means that their corresponding sides must be proportional. From this relationship, you can get an equation that involves only terms with x. Solve this equation and you should get the value for x that you showed.

5. Jun 4, 2009

fredrick08

Re: arcsin(1/x)=arctan(x)=>x=?

sorry i dont quite understand.... sin(a)=1/x so its triangle will have opposite=1 and hypotenuse=x and tan(b)=x so its triangle will have opposite=x and adjacent=1??? then from there how do i get a relationship with x's in it??

6. Jun 4, 2009

fredrick08

Re: arcsin(1/x)=arctan(x)=>x=?

i tried pythagoras.. all i get is x=x??? please someone can explain this to me?

7. Jun 4, 2009

fredrick08

Re: arcsin(1/x)=arctan(x)=>x=?

is get x=root(x^2+1)/x please can someone tell me how that simplifies to 1/2*root(2+root(20))???

8. Jun 4, 2009

HallsofIvy

Staff Emeritus
Re: arcsin(1/x)=arctan(x)=>x=?

That was on a different thread (and solved there).

9. Jun 4, 2009

fredrick08

Re: arcsin(1/x)=arctan(x)=>x=?

10. Jun 4, 2009

Cyosis

Re: arcsin(1/x)=arctan(x)=>x=?

You posted the same question in the advanced physics forum where I replied. However this does seem to be more fit for the calculus forum so I'll copy paste my reply here.

If you want to use identities you can take the sin on both sides and then use $\cos^2x+\sin^2x=1,\;1+\tan^2x=\sec^2x$ to simplify the right hand side.

11. Jun 4, 2009

JG89

Re: arcsin(1/x)=arctan(x)=>x=?

EDIT: Nevermind

12. Jun 4, 2009

Cyosis

Re: arcsin(1/x)=arctan(x)=>x=?

Arcsin is injective on the domain [-1,1], where it is the inverse of the sine with domain [-pi/2,pi/2]. However here we don't have arcsin(a)=arcsin(b), but arcsin(a)=arcTAN(b).

Last edited: Jun 4, 2009