Alternate form of $c\cdot \cos(a+b)$, with $c$ inside the $\cos$? I need to find an alternative form of $c\cdot\cos(a+b)$.
The goal is to get $c$ into the $\cos$ (e.g. $\cos(a\cdot c+b\cdot c)$).
Can anyone help me with this problem or knows where i can look further?
Thanks in advance
thorb3n
Edit1: Sorry, i thínk i wrote it the wrong way. i mean $c\cdot \cos(a+b)$
 A: If I understand your question correctly, you want an equation of the form $c \cdot \cos(a+b) = \cos(f(a,b,c))$.
However, no such real-valued function $f$ exists. If such an equation above does exist, then it must also work for my favourite numbers $a=0$, $b=0$ and $c=10$. 
However, the left side of the equation is $10$, which is outside of the range of $\cos$, so no such function could exist. 
A: Starting with $c\cdot\cos(a+b)$, for most purposes I can think of, we wouldn't need to change how it's expressed. It's quite simple the way its currently expressed, and neatly tells you the range of output values ($-c$ to $+c$) and the roots of the expression ($a+b=n\pi$).
But if you do want to express it with everything inside the $\cos(\cdot)$, you could try setting the expression as $c\cdot\cos(a+b)=\cos(x)$, and finding $x=\cos^{-1}\left[c\cdot\cos(a+b)\right]$.
We can find approximations for $\cos^{-1}(z)$, such as $\cos^{-1}(z)\approx \frac\pi2-z-\frac{z^3}{6}$, for $-1<z<1$. We can then set $z=\cos{-1}\left[c\cdot\cos(a+b)\right]$.
This implies that: $$c\cdot\cos(a+b)\approx \cos\bigg(\frac{\pi}{2}-c\cdot\cos(a+b)-\frac{c^3}{6}\cos^3(a+b)\bigg)$$
However, this approximation only works for $-1<c<1$. But, as mdave16 points out, $\cos(x)$ is bounded as $-1<\cos(x)<1$, so not all values of $c$ are possible anyway (unless $x$ can be a complex number).

The figure above shows what the approximation looks like in the worst case, for $a=-1.8,c=1$, as we vary $b$. It turns out that $a$ and $b$ don't affect the maximum error but even when $c$ is close to $1$, the absolute error is at most $0.08$.
