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

[fredrick08]

Homework Statement

can someone please tell me why arcsin(1/x)=arctan(x)=>x=-\sqrt{}(-2+\sqrt{}20)/(-1+\sqrt{}5)?

 

[fredrick08]

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

 

[fredrick08]

nope... that doesn't work...

 

[Mark44]

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.

 

[fredrick08]

sorry i don't 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??

 

[fredrick08]

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

 

[fredrick08]

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

 

[HallsofIvy]

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

 

[fredrick08]

sorry forgot about that one.

 

[Cyosis]

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.

 

[JG89]

EDIT: Nevermind

 

[Cyosis]

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).