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 $c\cdot \cos(a+b)$

• $\cos(ab)$ is not related to $\cos a$, $\sin a$, $\sin b$ and $\cos b$ in a sensible way. – user228113 Oct 13 '17 at 8:19
• Welcome to MSE. Please use MathJax. – José Carlos Santos Oct 13 '17 at 8:26
• Why exactly do you want such a form? (I just want more background), and what have you tried? – mdave16 Oct 13 '17 at 10:48

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.

• Good answer. Although, no such real valued function could exist, since $\cos(\cdot)$ can be unbounded when it's allowed to take complex values :) – Jam Oct 13 '17 at 10:58
• @Jam True, I overlooked it thinking that OP might not be thinking about them. Great answer btw +1 – mdave16 Oct 13 '17 at 11:14

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 $0.08$.

• We can use the Taylor series for $\cos(x)$, and some tricks with a sawtooth wave to add periodicity, and a square wave to get the sign right, to show that $\cos(x)\approx \frac12(-1)^{\lfloor \frac{x}{\pi}+\frac12\rfloor}\left(2-(x-\pi(1+\lfloor \frac{x}\pi-\frac12\rfloor))^2\right)$, with maximum error of $0.23$. Then $c\cdot \cos(a+b)\approx \cos\left(\frac\pi2-\frac{c}2(-1)^{\lfloor\frac{a+b}{\pi}+\frac12\rfloor}\left(2-(a+b-\pi(1+\lfloor \frac{a+b}{\pi}-\frac12\rfloor))^2\right)\right)$, with maximum error of $0.23$, when $c=1$. The function $\lfloor\cdot\rfloor$ is the floor function. – Jam Oct 13 '17 at 13:41