Find the value of 2.arccos 3/√13 + arccot 16/63 + arccos 7/25
Answer: arctan 3/4
How to solve this?
Hello,
I've seen some problems solved this way, but I just can't do them...
I'll try, but MY way, that is to say : calculations !
Here are the formula I'll use :
$\displaystyle (1) ~:~ \arctan(z)=\text{arccot} \left(\tfrac 1z\right)$ (formula 18 : Inverse Tangent -- from Wolfram MathWorld)
$\displaystyle (2) ~:~ \arccos(z)=\tfrac \pi 2-\arctan \left(\tfrac{z}{\sqrt{1-z^2}}\right)$ (formula 12 : Inverse Cosine -- from Wolfram MathWorld)
$\displaystyle (3) ~:~ \arctan(x)\pm \arctan(y)=\arctan \left(\frac{x\pm y}{1 \mp xy}\right)$ (formula 36 : Inverse Tangent -- from Wolfram MathWorld)
Using formula (1), we have to solve :
$\displaystyle 2 \arccos \left(\tfrac{3}{\sqrt{13}}\right)+\arctan \left(\tfrac{63}{16}\right)+\arccos\left(\tfrac{7} {25}\right)$
Using formula (2) and after some simplifications, we have to solve :
$\displaystyle \pi -2 \arctan \left(\tfrac 32\right)+\arctan \left(\tfrac{63}{16}\right)+ \tfrac \pi 2-\arctan \left(\tfrac{7}{24}\right)$
$\displaystyle =\tfrac{3 \pi}{2}+\left(\arctan \left(\tfrac{63}{16}\right)-\arctan \left(\tfrac 32\right)\right)-\left(\arctan \left(\tfrac 32\right)+\arctan \left(\tfrac{7}{24}\right)\right)$
Using formula (3), we get :
$\displaystyle \tfrac{3 \pi}{2}+\arctan \left(\tfrac{6}{17}\right)-\arctan \left(\tfrac{86}{27}\right)$
Using formula (3) once again, we get :
$\displaystyle \tfrac{3 \pi}{2}+\arctan \left(-\tfrac 34\right)=\tfrac{3 \pi}{2}-\arctan \left(\tfrac 34\right)$
And I must say I'm lost here...