How Can You Prove arctan(1/v) Equals (π/2) - arctan(v)?

[Calixto]

how can I show that ... arctan(1/v) = (π/2) - arctan(v) ?

 

[jhicks]

Kind of a hint: v = v/1 = 1/(1/v). You just need to remember the definition of tan

 

[Calixto]

I'm sorry, I don't understand what you did with the v. And what about the pi/2 part?

 

[jhicks]

In words, arctan(x) is the angle formed by a right triangle whose side opposite the angle and adjacent sides have ratio x/1. What is the measure of the other non-right angle in this hypothetical triangle, and what would the tangent of this other angle be?

 

[Calixto]

Here's how I think of it... I rearranged the equation so that arctan(1/v) + arctan(v) = π/2

And I understand how the tan of, say ø = 1/v, making the tan of, say ß = v. That part makes sense. The part I'm getting confused on is the π/2, and what that does to the equation.

 

[jhicks]

what's the sum of the angles in a triangle? Trig functions assume that one angle is 90 degrees. If one angle is $\theta$, then what's the other? Draw a triangle to aid you in this. Make one such that $arctan(v)=\theta$. What's the other angle? What's the tangent of this other angle?

 

[Calixto]

Ok, the other angle must be 180 - (90+ø), correct? And the tangent of this angle would be v... right?

I'm still not seeing this proving business.

 

[jhicks]

And the tangent of this angle would be v... right?

No! For the other angle, which is the opposite side and which is the adjacent?