Find the Solution for arccot(1/cot(pi/5)) without Cyclometric Functions

[Hannisch]

Homework Statement

Calculate arccot \left( \frac{1}{cot( \pi/5)} \right) The answer may not contain any cyclometric functions.

Homework Equations

The Attempt at a Solution

Can someone tell me where I went wrong? Cause I'm going insaaaane over this problem!

arccot \left( \frac{1}{cot( \pi/5)} \right)

arccot(x) = \frac{arccos(x)}{arcsin(x)} = \frac{1}{arctan(x)}


\frac{1}{arctan\left( \frac{1}{cot( \pi/5)} \right)}


= \frac{1}{arctan\left( \frac{1}{ \frac{cos(\pi/5)}{sin(\pi/5)}} \right)}


= \frac{1}{arctan\left( \frac{sin(\pi/5)}{ cos(\pi/5)} \right)}


= \frac{1}{arctan\left(tan(\pi/5) \right)}


= \frac{1}{\pi/5} = \frac{5}{\pi}


And according to the practise test I'm doing, this is wrong.. help?

 

[Avodyne]

{\rm arccot}(x) \ne \frac{\arccos(x)}{\arcsin(x)}

Try drawing a triangle.

 

[Hannisch]

Ah that sucks.. why didn't my teacher say so? Is it true for arctan? cos I used arctanx = arcsinx / arccosx in class today (at the blackboard) and got the right answer and he didn't say anything?

Anyway, I solved the problem (thanks for that tip, I never think of using triangles!) by saying that arccot(1/cotx) = arccot(tanx), but tanx=cot(90-x), i.e. 3π/10. Thanks!

 

[Hannisch]

Actually, scratch that ^. I do get the right answer, but I didn't use that with tan, sorry.. I confused it, the other way around (tan(arccosx)= sin(arccosx)/cos(arccosx). My bad, weird/stressful day.

 

[Avodyne]

I solved the problem ... by saying that arccot(1/cotx) = arccot(tanx), but tanx=cot(90-x), i.e. 3π/10.

Correct! And as you've figured out, the ratios like tan=sin/cos apply to the trig functions, and not to their inverses.