how can I show that ... arctan(1/v) = (π/2) - arctan(v) ???
Kind of a hint: v = v/1 = 1/(1/v). You just need to remember the definition of tan
I'm sorry, I don't understand what you did with the v. And what about the pi/2 part?
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?
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.
what's the sum of the angles in a triangle? Trig functions assume that one angle is 90 degrees. If one angle is [itex]\theta[/itex], then what's the other? Draw a triangle to aid you in this. Make one such that [itex]arctan(v)=\theta[/itex]. What's the other angle? What's the tangent of this other angle?
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.
No! For the other angle, which is the opposite side and which is the adjacent?
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