Dolores tried to calculate the value of the following expression without using a calculator:
$$\frac{1+\cos 15^{\circ}\cdot\sin^2 37^{\circ}\,30^{'}-\sin^2 15^{\circ}-\cos 15^{\circ}\cdot\cos^2 37^{\circ}\,30^{'}}{\cos 15^{\circ}}$$
She solved this task in the following steps:
\begin{aligned}
&\frac{1+\cos 15^{\circ}\cdot\sin^2 37^{\circ}\,30^{'}-\sin^2 15^{\circ}-\cos 15^{\circ}\cdot\cos^2 37^{\circ}\,30^{'}}{\cos 15^{\circ}}\stackrel{(1)}=\cr
\stackrel{(1)}=&\frac{(1-\sin^2 15^{\circ})+\cos15^{\circ}\cdot\left(\sin^2 \,37^{\circ}\,30^{'}-\cos^2 \,37^{\circ}\,30^{'}\right)}{\cos15^{\circ}}\stackrel{(2)}=\cr
\stackrel{(2)}=&\frac{\cos^2 15^{\circ}-\cos15^{\circ}\cdot\cos75^{\circ}}{\cos 15^{\circ}}\stackrel{(3)}=\cr
\stackrel{(3)}=&\cos15^{\circ}-\cos 75^{\circ}\stackrel{(4)}=\cr
\stackrel{(4)}=&-2 \sin45^{\circ}\sin30^{\circ}\stackrel{(5)}=\cr
\stackrel{(5)}=&-2\frac{\sqrt2}{2}\cdot\frac12=-\frac{\sqrt2}{2}
\end{aligned}
In which step of her solution did Dolores make a mistake?
The mistake is in step (2). The correct simplification should be: $$ \frac{(1-\sin^215^{\circ})+\cos15^{\circ}\cdot\left(\sin^2 37^{\circ}\,30^{'}-\cos^2 37^{\circ}\,30^{'}\right)}{\cos15^{\circ}}=\frac{\cos^215^{\circ}+\cos15^{\circ}\cdot\cos75^{\circ}}{\cos15^{\circ}}. $$
The mistake is in step (3). The equality $$\frac{\cos^215^{\circ}-\cos15^{\circ}\cdot\cos75^{\circ}}{\cos15^{\circ}}=\cos15^{\circ}-\cos75^{\circ}$$ does not hold for $\cos15^{\circ}=0$. The fraction cannot be simplified without considering this condition.
The mistake is in step (4). She used an incorrect trigonometric identity. The correct modification should be: \begin{aligned} \cos15^{\circ}-\cos75^{\circ}&=-2\sin\frac{15^{\circ}+75^{\circ}}{2}\sin\frac{15^{\circ}-75^{\circ}}{2}\cr &=-2\sin45^{\circ}\sin(-30^{\circ})=2\sin45^{\circ}\sin30^{\circ} \end{aligned}
The mistake is in step (4). She used an incorrect trigonometric identity. The correct modification should be: \begin{aligned} \cos15^{\circ}-\cos75^{\circ}&=2\sin\frac{15^{\circ}+75^{\circ}}{2}\cos\frac{15^{\circ}-75^{\circ}}{2}=\cr &=2\sin45^{\circ}\cos(-30^{\circ})=2\sin45^{\circ}\cos30^{\circ} \end{aligned}
The mistake is in step (5). The correct simplification should be: $$-2\sin45^{\circ}\sin30^{\circ}=-2\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}=-\frac{\sqrt6}{2}$$
Dolores made a mistake in step (4). She should have used the correct trigonometric sum-to-product identity: $$\cos\alpha-\cos \beta=-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$ The correct solution is: \begin{aligned} &\frac{1+\cos 15^{\circ}\cdot\sin^2 37^{\circ}\,30^{'}-\sin^2 15^{\circ}-\cos 15^{\circ}\cdot\cos^2 37^{\circ}\,30^{'}}{\cos 15^{\circ}}=\cr =&\frac{(1-\sin^2 15^{\circ})+\cos15^{\circ}\cdot\left(\sin^2 \,37^{\circ}\,30^{'}-cos^2 \,37^{\circ}\,30^{'}\right)}{\cos15^{\circ}}=\cr =&\frac{\cos^2 15^{\circ}-\cos15^{\circ}\cdot\cos75^{\circ}}{\cos 15^{\circ}} =\cos15^{\circ}-\cos 75^{\circ}=\cr =&-2 \sin45^{\circ}\sin\left(-30^{\circ}\right)=2\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt2}{2} \end{aligned}