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

Answer: arctan 3/4

How to solve this?

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- Sep 24th 2008, 07:21 AMfardeen_genFind the arccos value?
Find the value of 2.arccos 3/√13 + arccot 16/63 + arccos 7/25

Answer: arctan 3/4

How to solve this? - Sep 25th 2008, 05:02 AMfardeen_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.

- Sep 25th 2008, 12:04 PMMoo
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 ! (Rofl)

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... :(