Emma and Francis tried to calculate the value of $\sin375^\circ$ without using a calculator.
Identify who made a mistake and in which step of their solution?
Emma´s solution:
$$\sin375^\circ \stackrel{(1)}=\sin\left(360^\circ+15^\circ\right)\stackrel{(2)}=\sin15^\circ\stackrel{(3)}=\sin\frac{30^\circ}{2}\stackrel{(4)}=\frac{\frac12}{2}\stackrel{(5)}=\frac14$$
Francis’s solution:
$$\sin375^\circ\stackrel{(1)}=\sin\left(360^\circ+60^\circ-45^\circ\right)\stackrel{(2)}=\sin\left(60^\circ-45^\circ\right)\stackrel{(3)}=\sin60^\circ-\sin45^\circ\stackrel{(4)}=\frac{\sqrt3-\sqrt2}{2}$$
They both made a mistake in step (2): $$\sin\left(360^\circ+15^\circ\right)\neq \sin15^\circ$$ and $$\sin\left(360^\circ+60^\circ-45^\circ\right) \neq\left(\sin60^\circ-45^\circ\right).$$
Only Emma made a mistake. The mistake is in step (4): $$\sin\frac{30^\circ}{2}\neq\frac{\sin30^\circ}{2}$$
Only Francis made a mistake. The mistake is in step (3): $$\sin\left(60^\circ-45^\circ\right)\neq\sin60^\circ-\sin45^\circ$$
They both made a mistake.
Emma made a mistake in step (4): $$\sin\frac{30^\circ}{2}\neq\frac{\sin30^\circ}{2}$$ Francis made a mistake in step (3): $$\sin\left(60^\circ-45^\circ\right)\neq\sin60^\circ-\sin45^\circ$$
Let’s review Emma’s corrected solution: $$\sin375^\circ=\sin\left(360^\circ+15^\circ\right)$$ We use the fact that $\sin\left(\alpha\right)$ is a periodic function, so $\sin\left(360^\circ+\alpha\right)=\sin\left(\alpha\right)$ for every angle $\alpha$. Therefore: $$\sin\left(360^\circ+15^\circ\right)=\sin 15^\circ=\frac{\sin30^\circ}{2}.$$ Next, using the half-angle identity $\left|\sin\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{2}}$, we obtain: $$\sin\frac{30^\circ}{2}=\sqrt{\frac{1-\cos30^\circ}{2}}=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\frac{\sqrt{2-\sqrt3}}{2}$$
Now, let’s review Francis’s corrected solution: $$\sin375^\circ=\sin\left(360^\circ+60^\circ-45^\circ\right)=\sin\left(60^\circ-45^\circ\right)$$
Next, using the trigonometric identity $\sin\left(\alpha-\beta\right)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$, we obtain: $$\sin\left(60^\circ-45^\circ\right)=\sin60^\circ\cos45^\circ-\cos60^\circ\sin45^\circ=\frac{\sqrt3}{2}\frac{\sqrt2}{2}-\frac12\frac{\sqrt2}{2}=\frac{\sqrt6-\sqrt2}{4}.$$ Finally, note that: \begin{aligned} \frac{\sqrt{2-\sqrt3}}{2}&=\frac{2\sqrt{2-\sqrt3}}{4}=\frac{\sqrt{8-\sqrt{48}}}{4}=\frac{\sqrt{6-\sqrt{4\cdot2\cdot6}+2}}{4}=\cr &=\frac{\sqrt{6-2\cdot\sqrt2\cdot\sqrt6+2}}{4}=\frac{\sqrt{\left(\sqrt6-\sqrt2\right)^2}}{4}=\frac{\sqrt6-\sqrt2}{4}\end{aligned}