Answer
The exact value of the Mach speed of the aircraft is $\underline{M=4\sqrt{2-\sqrt{3}}+2\sqrt{3}\sqrt{2-\sqrt{3}}}$ and the approximate value is $\underline{M\approx 3.9.}$
Work Step by Step
We know that the relationship between the cone’s vertex angle, $\theta,$ and the Mach speed, $M,$ of an aircraft that is flying faster than the speed of sound is:
$\sin \frac{\theta }{2}=\frac{1}{M}.$
We use the half angle formula for the half angle of sine in terms of cosine. That means
$\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$
Since, Mach speed is positive, take only the positive value. That means,
$\begin{align}
& \frac{1}{M}=\sin \frac{\theta }{2} \\
& =+\sqrt{\frac{1-\cos \theta }{2}}
\end{align}$
Since, the provided value of the cone’s vertex angle, $\theta =\frac{\pi }{6}.$
$\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}}=\sqrt{\frac{1-\cos \frac{\pi }{6}}{2}}$
Since, $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}.$
$\sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \frac{\pi }{6}}{2}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}$
$\text{That means at }\theta =\frac{\pi }{6},$
$\begin{align}
& \sin \frac{\theta }{2}=\frac{\sqrt{2-\sqrt{3}}}{2} \\
& \frac{1}{M}=\frac{\sqrt{2-\sqrt{3}}}{2} \\
\end{align}$
And take the reciprocal on both sides. That means:
$M=\frac{2}{\sqrt{2-\sqrt{3}}}$
Then, divide the numerator and denominator of the right hand side $\sqrt{2-\sqrt{3}}.$ Then,
$M=\frac{2}{\sqrt{2-\sqrt{3}}}.\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}=\frac{2\sqrt{2-\sqrt{3}}}{2-\sqrt{3}}$
Divide the numerator and denominator of the right hand side $2+\sqrt{3}.$ Then,
$M=\frac{2\sqrt{2-\sqrt{3}}}{2-\sqrt{3}}.\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{4\sqrt{2-\sqrt{3}}+2\sqrt{3}\sqrt{2-\sqrt{3}}}{4-3}=4\sqrt{2-\sqrt{3}}+2\sqrt{3}\sqrt{2-\sqrt{3}}$
Thus, the approximate value of $M$ as a decimal to the nearest tenth calculated using a calculator is:
$M=4\sqrt{2-\sqrt{3}}+2\sqrt{3}\sqrt{2-\sqrt{3}}\approx 3.9$.
