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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?.

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

3 Answers 3

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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.

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

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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.

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