Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 81

The exact value of the Mach speed of the aircraft is $\underline{M=\sqrt{2-\sqrt{2}}.\left( 2+\sqrt{2} \right)}$ and the approximate value is $\underline{M\approx 2.6.}$

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 the 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 given value of the cone’s vertex angle $\theta =\frac{\pi }{4}.$ $\begin{align} & \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{2}} \\ & =\sqrt{\frac{1-\cos \frac{\pi }{4}}{2}} \end{align}$ Since, $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\text{ or }\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$, $\begin{align} & \sin \frac{\theta }{2}=\sqrt{\frac{1-\cos \frac{\pi }{4}}{2}} \\ & =\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}} \end{align}$ Solving the equation as follows: $\begin{align} & \sin \frac{\theta }{2}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} \\ & =\sqrt{\frac{2-\sqrt{2}}{4}} \\ & =\frac{\sqrt{2-\sqrt{2}}}{2} \end{align}$ $\text{That means at }\theta =\frac{\pi }{4},$ $\begin{align} & \sin \frac{\theta }{2}=\frac{\sqrt{2-\sqrt{2}}}{2} \\ & \frac{1}{M}=\frac{\sqrt{2-\sqrt{2}}}{2} \\ \end{align}$ And take the reciprocal on both sides. That means $M=\frac{2}{\sqrt{2-\sqrt{2}}}$ And divide the numerator and the denominator of the right-hand side $\sqrt{2-\sqrt{2}}.$ Then, we have $\begin{align} & M=\frac{2}{\sqrt{2-\sqrt{2}}}.\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}} \\ & =\frac{2\sqrt{2-\sqrt{2}}}{2-\sqrt{2}} \end{align}$ Divide the numerator and denominator of the right-hand side $2+\sqrt{2}.$ Then, we have $\begin{align} & M=\frac{2\sqrt{2-\sqrt{2}}}{2-\sqrt{2}}.\frac{2+\sqrt{2}}{2+\sqrt{2}} \\ & =\frac{4\sqrt{2-\sqrt{2}}+2\sqrt{2}\sqrt{2-\sqrt{2}}}{4-2} \\ & =\frac{2\left( 2\sqrt{2-\sqrt{2}}+\sqrt{2}\sqrt{2-\sqrt{2}} \right)}{2} \\ & =2\sqrt{2-\sqrt{2}}+\sqrt{2}\sqrt{2-\sqrt{2}} \end{align}$ Also, taking $\sqrt{2-\sqrt{2}}$ common, we have $M=\sqrt{2-\sqrt{2}}.\left( 2+\sqrt{2} \right)$ Thus, the approximate value of $M$ as a decimal to the nearest tenth calculated using a calculator is $\begin{align} & M=\sqrt{2-\sqrt{2}}.\left( 2+\sqrt{2} \right) \\ & \approx 2.6 \end{align}$