How to simplify $\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$? The problem is as follows:
Simplify the following expression:
$B=\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&7.5\\
2.&6\\
3.&8\\
4.&5\\
\end{array}$
I'm not sure how to proceed here because the division seems kind of complicated to simplify.
But I could spot that suspicioulsy $\sqrt{2}= \csc 45^\circ$
and also $2= \csc 30^\circ$
But I don't know if these would come into play in solving this problem as it is challenging. Perhaps does it exist a way to solve this without much fuss?.
I could also spot that:
$\sin \left(90-\frac{45}{2}\right)= \cos \frac{45}{2}$
$\sin 55^\circ = \cos 35^\circ$
and
$\sin 57.5^\circ= \cos 32.5^\circ = \sin \frac{65}{2}^\circ$
The rest I presume that involves the simplifcation of the expresion using sum to product formulas. But I got stuck with those. Can someone help me here?.

The denominator is $\sin(55^\circ) \sin(57.5^\circ) \sin(67.5^\circ)$ and note that the angles add up to $180^\circ$.
Using the identity $\sin \alpha \cdot \sin \beta \cdot \sin \gamma = \frac14(\sin 2\alpha + \sin 2 \beta + \sin 2\gamma)$ listed in Wikipedia Further "conditional" identities for the case α + β + γ = 180° that is only valid if the angles sum to $180^\circ$, the denominator becomes
$$\frac14(\sin(110^\circ)+\sin(115^\circ)+\sin(135^\circ)) = \frac14(\sin(70^\circ)+\sin(65^\circ)+\sin(45^\circ))$$
since $\sin(180^\circ-\theta)=\sin(\theta)$. Then use the value for $\sin45^\circ$ and the fact that $\sin(90^\circ-\theta)=\cos(\theta)$ to obtain
$$\frac{1}{4}(\frac{1}{\sqrt2}+\cos(20^\circ)+\cos(25^\circ))$$
for the denominator. Writing the entire fraction,
$$\frac{\sqrt{2}+2\cos(20^\circ)+2\cos(25^\circ)}{\frac{1}{4}(\frac{1}{\sqrt2}+\cos(20^\circ)+\cos(25^\circ))} = \frac{8(\sqrt{2}+2\cos(20^\circ)+2\cos(25^\circ))}{2(\frac{1}{\sqrt2}+\cos(20^\circ)+\cos(25^\circ))} = 8$$
is the desired result.
Edited to add:
Here is a quick proof of the identity mentioned previously.
$$\sin\alpha\sin\beta\sin\gamma = \sin\alpha\sin\beta\sin(180^\circ-(\alpha+\beta))=\sin\alpha\sin\beta\sin(\alpha+\beta)$$
since $\alpha+\beta+\gamma=180^\circ$ and $\sin(180^\circ-\theta)=\sin(\theta)$. Use the angle sum identity for sine to obtain
$$\sin\alpha\sin\beta(\sin\alpha\cos\beta+\cos\alpha\sin\beta) = \sin^2\alpha\sin\beta\cos\beta+\sin^2\beta\sin\alpha\cos\alpha$$
which is transformed using the double- and half-angle formulas for sine
$$\frac12(\sin^2\alpha\sin2\beta+\sin^2\beta\sin2\alpha) = \frac14((1-\cos2\alpha)\sin2\beta+(1-\cos2\beta)\sin2\alpha)$$
$$=\frac14(\sin2\alpha+\sin2\beta-(\cos2\alpha\sin2\beta+\cos2\beta\sin2\alpha))=\frac14(\sin2\alpha+\sin2\beta-\sin(2\alpha+2\beta))$$
which is the desired result if the last term is equal to $\sin2\gamma$. That is easily shown since
$$\sin2\gamma=\sin(2(180^\circ-(\alpha+\beta)))=\sin(360^\circ-2(\alpha+\beta))$$
$$=\sin360^\circ\cos(2(\alpha+\beta))-\cos(360^\circ)\sin(2(\alpha+\beta)) = -\sin(2\alpha+2\beta)$$
completes the proof.
