Olympiad Trigonometric Problem (Continued)

 Previously we asked you, the educators or yourself, if you are a student to think of way(s) to find the ratio of $\dfrac{PR}{QR}$ in this thread (Olympiad Trigonometric Problem), we will now lead you to a credible way to solve for that intriguing hard trigonometric problem.

We in fact, should have the eagle eye sight and should be able to tell offhand that we suspect $\angle P=\angle Q$. 

The effort hence should focus on how to prove that is the case and hence deduce that $\dfrac{PR}{QR}=1$.

This is when inequality comes into the picture and becomes the greatest helper to solve for intriguing problem!

Suppose $\angle Q>\angle P$. Then regardless if $Q$ is an obtuse angle, the half of these $\angle P$ and $\angle Q$ are less than $90^{\circ}$and so we have

$\cos \left(\dfrac{P}{2}\right)>\cos \left(\dfrac{Q}{2}\right)$

$\implies \cos^{48} \left(\dfrac{P}{2}\right)>\cos^{48} \left(\dfrac{Q}{2}\right)---(*)$

and we also have

$\sin \left(\dfrac{Q}{2}\right)>\sin \left(\dfrac{P}{2}\right)$

$\implies \sin^{23} \left(\dfrac{Q}{2}\right)>\sin^{23} \left(\dfrac{P}{2}\right)---(**)$

Multiply the two inequalities (*) and (**) together, we see that

$\cos^{48} \left(\dfrac{P}{2}\right)\sin^{23} \left(\dfrac{Q}{2}\right)>\cos^{48} \left(\dfrac{Q}{2}\right)\sin^{23} \left(\dfrac{P}{2}\right)$

and this contradicts to what we are told that the above is an equality. So, $\angle P=\angle Q$ and hence, $PR=QR$, or more specifically, $\dfrac{PR}{QR}=1$

The question now remains, can we prove $\angle P=\angle Q$ based on the given equality using only the trigonometric method? We will try it in the next blog post...so please stay tuned!