Find the arccos value?

Hello,

Quote:

Originally Posted by fardeen_gen

I was told that assuming an inverse trigonometric function as an angle simplifies the problem. This doesn't seem to help me in this problem.

I've seen some problems solved this way, but I just can't do them...

Quote:

Originally Posted by fardeen_gen

Find the value of 2.arccos 3/√13 + arccot 16/63 + arccos 7/25

Answer: arctan 3/4

How to solve this?

I'll try, but MY way, that is to say : calculations ! (Rofl)

Here are the formula I'll use :

$(1) ~:~ \arctan(z)=\text{arccot} \left(\tfrac 1z\right)$ (formula 18 : Inverse Tangent -- from Wolfram MathWorld)

$(2) ~:~ \arccos(z)=\tfrac \pi 2-\arctan \left(\tfrac{z}{\sqrt{1-z^2}}\right)$ (formula 12 : Inverse Cosine -- from Wolfram MathWorld)

$(3) ~:~ \arctan(x)\pm \arctan(y)=\arctan \left(\frac{x\pm y}{1 \mp xy}\right)$ (formula 36 : Inverse Tangent -- from Wolfram MathWorld)

Using formula (1), we have to solve :
$2 \arccos \left(\tfrac{3}{\sqrt{13}}\right)+\arctan \left(\tfrac{63}{16}\right)+\arccos\left(\tfrac{7} {25}\right)$

Using formula (2) and after some simplifications, we have to solve :
$\pi -2 \arctan \left(\tfrac 32\right)+\arctan \left(\tfrac{63}{16}\right)+ \tfrac \pi 2-\arctan \left(\tfrac{7}{24}\right)$

$=\tfrac{3 \pi}{2}+\left(\arctan \left(\tfrac{63}{16}\right)-\arctan \left(\tfrac 32\right)\right)-\left(\arctan \left(\tfrac 32\right)+\arctan \left(\tfrac{7}{24}\right)\right)$

Using formula (3), we get :

$\tfrac{3 \pi}{2}+\arctan \left(\tfrac{6}{17}\right)-\arctan \left(\tfrac{86}{27}\right)$

Using formula (3) once again, we get :

$\tfrac{3 \pi}{2}+\arctan \left(-\tfrac 34\right)=\tfrac{3 \pi}{2}-\arctan \left(\tfrac 34\right)$

And I must say I'm lost here... :(