# 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?.

• Presumably there are degree symbols missing in $90-\frac{45}2$? Feb 26, 2021 at 2:16

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.

Hint

$$\sqrt2=2\sin45^\circ=2\sin2\cdot\dfrac{45}2^\circ=?$$

Use https://mathworld.wolfram.com/ProsthaphaeresisFormulas.html for $$\cos20^\circ+\cos25^\circ$$

Finally $$\cos\dfrac52^\circ+\cos\left(90-\dfrac{45}2\right)^\circ=?$$

I think you may find some help here at Wikipedia. There are general trig identities and there are some... surprising identities. I've linked to a section with some identities that only hold when $$\alpha + \beta + \gamma = \pi = \frac{\tau}{2} = 180°$$, which if you look closely might (might!) help in the denominator.

In particular, $$\sin \alpha \cdot \sin \beta \cdot \sin \gamma = \frac14(\sin 2\alpha + \sin 2 \beta + \sin 2\gamma)$$

Earlier on the same page is a section of identities without variables, which might also help, though I haven't tried to fully solve based on either of these--just looking at resources.

Edit: Is there the possibility you're supposed to use small-angle identities to [cough] "ignore" some of the tiny leftover bits. $$\cos 2.5° = 0.999$$. It's not rigorous but it might nonetheless be intended behavior on the part of the instructor.