1. ## 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

2. 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 $\displaystyle \cos\left(\frac{\arccos(u)}{3}\right)$, then it doesnt simplify to anything too nice.

3. 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

4. $\displaystyle \cos\left(\frac{\arccos(x)}{3}\right)$

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

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

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

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

So rewriting we get

\displaystyle \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}

5. Mathstud28,

brian311