# arccos simplification

• Nov 20th 2008, 08:44 PM
brian311
arccos simplification
Hello everyone,

Can anyone tell me how the following could be arranged to cancel out both the arccos and the cos? In other words, how could I include the 3 in the arccos term so that the cos and cos^-1 would cancel each other out?

cos((cos^-1(u))/3)

Thanks
• Nov 20th 2008, 08:46 PM
Mathstud28
Quote:

Originally Posted by brian311
Hello everyone,

Can anyone tell me how the following could be arranged to cancel out both the arccos and the cos? In other words, how could I include the 3 in the arccos term so that the cos and cos^-1 would cancel each other out?

cos((cos^-1(u))/3)

Thanks

If this is indeed $\cos\left(\frac{\arccos(u)}{3}\right)$, then it doesnt simplify to anything too nice.
• Nov 20th 2008, 08:55 PM
brian311
Thanks for the quick reply Mathstud28,

Actually it is:

cos((cos^-1(B/A))/3)

where B and A are symbolic constants. Is there a way to cancel the cos^-1 and cos now?

Thanks
• Nov 20th 2008, 09:05 PM
Mathstud28
$\cos\left(\frac{\arccos(x)}{3}\right)$

Let $y=\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$

Let $e^{ix}=\varphi$ so we have that $y=\frac{\varphi+\frac{1}{\varphi}}{2}$

Solving and taking the appropriate (this is subjective...you figure out why) root gives

$\varphi=e^{ix}=\sqrt{y^2-1}+y\implies{x}=-i\ln\left(\sqrt{y^2-1}+y\right)=\arccos(y)$

So rewriting we get

\begin{aligned}\cos\left(\frac{\arccos(x)}{3}\righ t)&=\frac{e^{\frac{i\cdot{-i}\ln\left(\sqrt{x^2-1}+x\right)}{3}}+e^{\frac{-i\cdot{-i}\ln\left(\sqrt{x^2-1}+x\right)}{3}}}{2}\\
&=\frac{e^{\frac{-1}{3}\ln\left(\sqrt{x^2-1}+x\right)}+e^{\frac{1}{3}\ln\left(\sqrt{x^2-1}+x\right)}}{2}\\
&=\frac{\sqrt[-3]{\sqrt{x^2-1}+x}+\sqrt[3]{\sqrt{x^2-1}+x}}{2}\end{aligned}
• Nov 20th 2008, 09:08 PM
brian311
Mathstud28,