PDA

View Full Version : Interesting yet challenging proof...


Calixto
Mar24-08, 11:19 PM
how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???

jhicks
Mar24-08, 11:21 PM
Kind of a hint: v = v/1 = 1/(1/v). You just need to remember the definition of tan

Calixto
Mar24-08, 11:28 PM
I'm sorry, I don't understand what you did with the v. And what about the pi/2 part?

jhicks
Mar24-08, 11:34 PM
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
Mar24-08, 11:47 PM
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
Mar25-08, 12:24 AM
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
Mar25-08, 12:29 AM
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
Mar25-08, 01:15 AM
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?