Welcome to our community

Be a part of something great, join today!

Problem Of The Week #486 October 11th 2021

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,941
Here is this week's POTW:

-----

Evaluate $\dfrac{1}{\sin 6^{\circ}}+\dfrac{1}{\sin 78^{\circ}}-\dfrac{1}{\sin 42^{\circ}}-\dfrac{1}{\sin 66^{\circ}}$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,941
Congratulations to lfdahl for his correct solution (Cool) , which you can find below:
\[\frac{1}{\sin 6^{\circ}}+\frac{1}{\sin 78^{\circ}}-\frac{1}{\sin 42^{\circ}}-\frac{1}{\sin 66^{\circ}} \\\\\\=\frac{\sin 66^{\circ}-\sin 6^{\circ}}{\sin 6^{\circ} \sin 66^{\circ}}+\frac{\sin 42^{\circ}-\sin 78^{\circ}}{\sin 42^{\circ} \sin 78^{\circ}} \\\\\\= \frac{2\sin \left ( \frac{66^{\circ}-6^{\circ}}{2} \right )\cos \left ( \frac{66^{\circ}+6^{\circ}}{2}\right )}{\frac{1}{2}\left ( \cos (66^{\circ}-6^{\circ}) -\cos \left ( 66^{\circ}+6^{\circ}\right )\right )}+\frac{2\sin \left ( \frac{42^{\circ}-78^{\circ}}{2} \right )\cos \left ( \frac{42^{\circ}+78^{\circ}}{2}\right )}{\frac{1}{2}\left ( \cos (42^{\circ}-78^{\circ}) -\cos \left ( 42^{\circ}+78^{\circ}\right )\right )} \\\\\\=\frac{2 \sin 30^{\circ}\cos 36^{\circ}}{\frac{1}{2}\left ( \cos 60^{\circ}-\cos 72^{\circ} \right )}-\frac{2 \sin 18^{\circ}\cos 60^{\circ}}{\frac{1}{2}\left ( \cos 36^{\circ}-\cos 120^{\circ} \right )} \\\\\\=\underbrace{\frac{4\cos 36^{\circ}}{1-\cos(2\cdot 36^{\circ})}}_A\underbrace{-\frac{4\cos(2\cdot 36^{\circ})}{1+2\cos 36^{\circ}}}_B\]
Let $\alpha = \cos 36^{\circ}$:

\[A = \frac{4\alpha }{3-4\alpha ^2}, \;\;\; B = \frac{4-8\alpha ^2}{1+2\alpha }\]

It is not hard to show*, that: $\alpha = \frac{1+\sqrt{5}}{4}$. Inserting this in $A$ and $B$ yields:

\[A = \frac{2+2\sqrt{5}}{3-\sqrt{5}},\;\;\;B = \frac{2-2\sqrt{5}}{3+\sqrt{5}}\]

Finally, we get: \[A + B = 8.\]

(*). I use the well-known trick to calculate $\sin 18^{\circ}$ first:

Put $Y = 18^{\circ}$.

Then $\sin 2Y = \cos 3Y \rightarrow 2\sin Y\cos Y = 4\cos^3Y-3\cos Y \rightarrow \\\\ 2 \sin Y = 4(1-\sin^2Y) -3 \rightarrow 4\sin^2Y +2\sin Y-1 = 0$ - the solution of which is: $\sin Y = \sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.

With $\cos 36^{\circ} = 1 – 2 \sin^2 18^{\circ}$ we immediately have the result.
 
Status
Not open for further replies.