How Can You Express Cos(a-b) in Terms of m and n?

[anemone]

Here is this week's POTW:

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Given that $$\frac{\sin (x-a)}{\sin (x-b)}=m$$ and $$\frac{\cos (x-a)}{\cos (x-b)}=n$$ where $$0\lt x \lt \frac{\pi}{2}$$ and $m$ and $n$ are two positive real numbers.

Express $$\cos (a − b)$$ in terms of $m$ and $n$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. Ackbach
2. Theia
3. lfdahl
4. kaliprasad
5. Opalg

Solution from Ackbach:

Spoiler

Given that $\sin(x-a)=m \sin(x-b)$ and $\cos(x-a)=n \cos(x-b)$, we can rewrite these two equations as
\begin{align*}
\sin(x)\cos(a)-\cos(x)\sin(a)&=m\left[\sin(x)\cos(b)-\cos(x)\sin(b)\right] \\
\cos(x)\cos(a)+\sin(x)\sin(a)&=n\left[\cos(x)\cos(b)+\sin(x)\sin(b)\right].
\end{align*}
We view this system as equations in $\sin(x)$ and $\cos(x)$:
\begin{align*}
\sin(x)\left[\cos(a)-m\cos(b)\right]+\cos(x)\left[-\sin(a)+m\sin(b)\right]&=0 \\
\sin(x)\left[\sin(a)-n\sin(b)\right]+\cos(x)\left[\cos(a)-n\cos(b)\right]&=0.
\end{align*}
It is impossible for $\sin(x)=\cos(x)=0$, so this system must be degenerate, which forces the determinant to be zero:
$$(\cos(a)-m\cos(b))(\cos(a)-n\cos(b))-(-\sin(a)+m\sin(b))(\sin(a)-n\sin(b))=0,$$
or
$$\cos^2(a)-(m+n)\cos(a)\cos(b)+mn\cos^2(b)+\sin^2(a)-(m+n)\sin(a)\sin(b)+mn\sin^2(b)=0,$$
or
$$1+mn-(m+n)\left[\cos(a)\cos(b)+\sin(a)\sin(b)\right]=0,$$
and hence
$$\cos(a-b)=\frac{1+mn}{m+n}.$$
Since $m$ and $n$ are both positive, the denominator cannot be zero.

Solution from Opalg:

Spoiler

Write the equations as $$\sin(x-a) = m\sin(x-b), \qquad \cos(x-a) = n\cos(x-b).$$ Square both equations and add, to get $$1 = n^2\cos^2(x-b) + m^2\sin^2(x-b). \qquad(1)$$ Next, $$\begin{aligned}\cos(a-b) &= \cos\bigl((x-b) - (x-a)\bigr) \\ &= \cos(x-b)\cos(x-a) + \sin(x-b)\sin(x-a) \\ &= n\cos^2(x-b) + m\sin^2(x-b). \qquad(2) \end{aligned}$$ Now multiply (2) by $n$ and subtract (1), to get $$n\cos(a-b) - 1 = m(n-m)\sin^2(x-b). \qquad(3)$$ Similarly, multiply (2) by $m$ and subtract (1), to get $$m\cos(a-b) - 1 = n(m-n)\cos^2(x-b). \qquad(4)$$ If $m=n$ then it follows from (1) that $m=n= 1$ and then from (3) or (4) that $\cos(a-b) = 1.$

Otherwise, if $m\ne n$ then it follows from (3) and (4) that $$\frac{n\cos(a-b) - 1}{m(n-m)} + \frac{m\cos(a-b) - 1}{n(m-n)} = \sin^2(x-b) + \cos^2(x-b) = 1.$$ Therefore $$n(n\cos(a-b) - 1) - m(m\cos(a-b) - 1) = mn(n-m),$$ $$(n^2-m^2)\cos(a-b) = mn(n-m) + n - m,$$ $$(m+n)\cos(a-b) = mn+1$$ and finally $$\boxed{\cos(a-b) = \frac{mn+1}{m+n}}.$$ Notice that this last formula also covers the case when $m=n=1$, since it then gives the correct value $\cos(a-b) = 1.$