MHB Trigonometry Challenge II: Solving for $m,\,n$ and $A$

[anemone]

Find $m,\,n$ and $A$ such that $\sqrt{9-8\cos 40^{\circ}}=m+n\cos A^{\circ}$ where $m,\,n\in N$.

 

[Opalg]

Find $m,\,n$ and $A$ such that $\sqrt{9-8\cos 40^{\circ}}=m+n\cos A^{\circ}$ where $m,\,n\in N$.

I sort of cheated here by using a calculator to guess the answer, and then working backwards.

[sp]The formula $\sin3\theta = 3\sin\theta - 4\sin^3\theta$ shows that $\frac12= \sin30^\circ = 3\sin10^\circ - 4\sin^310^\circ$, so that $8\sin^310^\circ = 6\sin10^\circ - 1$. Then $$\begin{aligned} 9 - 8\cos40^\circ &= 9 - 8(1-2\sin^220^\circ) \\ &= 1 + 16\sin^220^\circ \\ &= 1 + 16(2\sin10^\circ\cos10^\circ)^2 \\ &= 1 + 64\sin^210^\circ(1-\sin^210^\circ) \\&= 1 + 64\sin^210^\circ - 64\sin^410^\circ \\&= 1 + 64\sin^210^\circ - 8\sin10^\circ(6\sin10^\circ - 1) \\ &= 1 + 8\sin10^\circ + 16\sin^210^\circ \\ &= (1+4\sin10^\circ)^2.\end{aligned}$$ Therefore $\sqrt{\mathstrut9 - 8\cos40^\circ} = 1+4\sin10^\circ = 1 + 4\cos80^\circ$. So $m=1,\ n=4,\ A=80$.[/sp]

 

[anemone]

Hi Opalg,

Thanks for your solution and thanks too for being honest to me. :)

A quite similar solution (of other) that I want to share with MHB:

$\begin{align*}\sqrt{9-8\cos 40^{\circ}}&=\sqrt{9-8(2\sin 30^{\circ}\cos 40^{\circ})}\\&=\sqrt{9-8(\sin 70^{\circ}-\sin 10^{\circ})}\\&=\sqrt{9-8(1-2\sin^2 10^{\circ}-\sin 10^{\circ} )}\\&=\sqrt{1+8\sin 10^{\circ}+16\sin^2 10^{\circ}}\\&=\sqrt{(1+4\sin 10^{\circ})^2}\\&=1+4\sin 10^{\circ}\\&=1+4\cos 80^{\circ}\end{align*}$

Therefore, $m=1,\,n=4$ and $A=80$.