Jtdylktuy

I'm trying to calculate the expression: $$frac{1}{sinfrac{pi}{15}}+frac{1}{sinfrac{2pi}{15}}-frac{1}{sinfrac{4pi}{15}}+frac{1}{sinfrac{8pi}{15}}$$ and show that it is equal $4sqrt{3}$.

I was trying to group the summands and calculate sums of $$frac{1}{sinfrac{pi}{15}}+frac{1}{sinfrac{2pi}{15}} hspace{0.5cm}text{and} hspace{0.5cm} -frac{1}{sinfrac{4pi}{15}}+frac{1}{sinfrac{8pi}{15}}$$ where we get $$frac{2cosfrac{2pi}{15}+1}{sinfrac{2pi}{15}}-frac{2cosfrac{4pi}{15}-1}{sinfrac{8pi}{15}}$$ but unfortunately this sum is not simplified.

How to prove this equality?

$$frac{1}{sinfrac{pi}{15}}+frac{1}{sinfrac{2pi}{15}}-frac{1}{sinfrac{4pi}{15}}+frac{1}{sinfrac{8pi}{15}}\ =frac{1}{sin12°}+frac{1}{sin24°}-frac{1}{sin48°}+frac{1}{sin96°} $$

We split them into two groups as shown:

$$frac{1}{sin12°}-frac{1}{sin48°} hspace{0.5cm}text{and}hspace{0.5cm} frac{1}{sin24°}+frac{1}{sin96°}$$
And we see that (from sum-to-product and product-to-sum)

$Large{frac{1}{sin12°}-frac{1}{sin48°}\ = frac{sin48°-sin12°}{sin12°sin48°}\ = frac{2cos30°sin18°}{{frac12}(cos36°-cos60°)}\=frac{2(frac{sqrt3}{2})sin18°}{{frac12}(cos36°-{frac12})}\=frac{4sqrt3sin18°}{2cos36°-1}}$

and

$Large{frac{1}{sin24°}+frac{1}{sin96°}\=frac{sin24°+sin96°}{sin24°sin96°}\=frac{2sin60°cos36°}{{frac12}(cos72°-cos120°)}\=frac{2({frac{sqrt3} 2})cos36°}{{frac12}(sin18°+{frac12})}\=frac{4sqrt3cos36°}{2sin18°+1}}$

Hence it remains to find

$$frac{4sqrt3sin18°}{2cos36°-1} + frac{4sqrt3cos36°}{2sin18°+1}$$

From here we can determine $2cos36°-1=2sin18° $ and $ 2sin18°+1=2cos36°$. By plugging in these into the denominators and simplifying we get $4sqrt3$, which is what we want. $ _square$

I use degrees:
$$frac{1}{sin 12^circ}+frac{1}{sin 24^circ}-frac{1}{sin48^circ}+frac{1}{sin96^circ}=\
frac{sin 96^circ+sin 12^circ}{sin 12^circsin 96^circ}+frac{sin 48^circ-sin 24^circ}{sin 24^circsin 48^circ} =\
frac{2sin 54^circcos 42^circ}{sin 12^circsin 96^circ}+frac{2cos 36^circsin 12^circ}{sin 24^circsin 48^circ} =\
frac{2cos 36^circrequire{cancel} cancel{sin 48^circ}}{2sin 12^circcancel{sin 48^circ}cos 48^circ}+frac{2cos 36^circcancel{sin 12^circ}}{2cancel{sin 12^circ}cos 12^circsin 48^circ} =\
cos 36^circcdot frac{cos 12^circsin 48^circ+sin 12^circcos 48^circ}{sin 12^circcos 12^circsin 48^circcos 48^circ} =\
frac{4cos 36^circsin 60^circ}{sin 24^circsin 96^circ}=\
frac{4cos 36^circsin 60^circ}{frac12(cos 72^circ-cos 120^circ)}=\
frac{4cos 36^circsin 60^circ}{frac12((cos 36^circ-frac12)+frac12)}=4sqrt{3}.
$$
You can see see here: $cos72^circ = cos36^circ-frac12$.

Well, this is not a full solution (I will not do the calculations), but I guess this is one of the ways to reach the end ...

1. $sin(frac{pi}{15}) = sin(frac{pi}{6} - frac{pi}{10})$


2. $sin(frac{pi}{10}) = sin(frac{1}{2}*frac{pi}{5})$


3. for the $frac{pi}{5}$ use the identity: $sin(5x) = 5sin(x)-20sin^3(x) + 16sin^5(x)$. It has a straightforward solution

Rest of the angles are possible to calculate with the double angle formula.

99,9% not indended to be solved this way, but ...

Hint:

Using $frac{1}{sin 8^circ}+frac{1}{sin 16^circ}+....+frac{1}{sin 4096^circ}+frac{1}{sin 8192^circ}=frac{1}{sin alpha}$,find $alpha$,

$$dfrac1{sin12^circ}+dfrac1{sin24^circ}+dfrac1{sin48^circ}+dfrac1{sin96^circ}-dfrac2{sin48^circ}$$

$$=cot6^circ-cot96^circ-dfrac2{sin48^circ}$$

$cot6^circ-cot96^circ=cot6^circ+tan 6^circ=dfrac2{sin12^circ}$

Now $dfrac1{sin12^circ}-dfrac1{sin48^circ}=dfrac{sin48^circ-sin12^circ}{sin48^circsin12^circ}=dfrac{4sin18^circcos30^circ}{cos36^circ-cos60^circ}$

Using Proving trigonometric equation $cos(36^circ) - cos(72^circ) = 1/2$, $cos36^circ-cos60^circ=sin18^circ$