Proof that two trig functions = Pi/2

• Feb 19th 2011, 08:13 PM
skyd171
Proof that two trig functions = Pi/2
The problem: prove that

$ArcTan(x)+ArcCot(x) = \Pi/2$

I am not sure how you "prove" this.
• Feb 19th 2011, 08:21 PM
mr fantastic
Quote:

Originally Posted by skyd171
The problem: prove that

$ArcTan(x)+ArcCot(x) = \Pi/2$

I am not sure how you "prove" this.

Define $f(x) = \arctan(x) + arccot(x)$. Show that f'(x) = 0. Therefore .... Now note that f(1) = .... Therefore ....
• Feb 19th 2011, 08:30 PM
skyd171
Thanks.
• Feb 19th 2011, 08:52 PM
Prove It
Quote:

Originally Posted by skyd171
The problem: prove that

$ArcTan(x)+ArcCot(x) = \Pi/2$

I am not sure how you "prove" this.

Actually, this is not true...

$\displaystyle \arctan{x} + \textrm{arccot}\,{x} = \frac{\pi}{2}$, not $\displaystyle \frac{\Pi}{2}$...
• Feb 20th 2011, 04:22 AM
HallsofIvy
Now that's going to confuse a lot of people, Prove It!

More generally, tan and cotan, sin and cos, sec and csc "swap" complementary angles. That is, if $\theta$ and $\phi$ are the two non-right angles in a right triangle, then [tex]tan(\theta)= cot(\phi), $sin(\theta)= cos(\phi)$, [tex], and $sec(\theta)= csc(\phi)$. And, of course, those two angles add to $\frac{\pi}{2}$ (did I get it right, Prove It?).
• Feb 20th 2011, 04:35 AM
Prove It
Quote:

Originally Posted by HallsofIvy
Now that's going to confuse a lot of people, Prove It!

More generally, tan and cotan, sin and cos, sec and csc "swap" complementary angles. That is, if $\theta$ and $\phi$ are the two non-right angles in a right triangle, then [tex]tan(\theta)= cot(\phi), $sin(\theta)= cos(\phi)$, [tex], and $sec(\theta)= csc(\phi)$. And, of course, those two angles add to $\frac{\pi}{2}$ (did I get it right, Prove It?).

I just meant because $\displaystyle \pi$ is the ratio of the circumference to the diameter of a circle, while $\displaystyle \Pi$ is the symbol for product :P