MHB Can You Solve the Triangle $PQR$ Equation from POTW #238?

[anemone]

Here is this week's POTW:

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In a triangle $PQR$ with its sides $p,\,q$ and $r$ where $2\angle P=3\angle Q$, prove that $(p^2-q^2)(p^2+pr-q^2)=q^2r^2$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution!:)

Here's the suggested model solution:

Spoiler

By the Law of Sines, the targeted identity can be rewritten as follows:

$$\left(\sin^2P-\sin^2 Q\right)\left(\sin^2 P+\sin P \sin R -\sin^2 Q\right)=\sin^2 Q \sin^2 R$$(*)

Note that

$$\cos 2P - \cos 2Q =-2\sin(P+Q) \sin (P-Q)$$

$$1-2\sin^2 P -\left(1-2\sin^2 Q\right)=-2\sin(P+Q) \sin (P-Q)$$

$$\sin^2 P -\sin^2 Q=\color{yellow}\bbox[5px,green]{\sin (P+Q)}\color{black} \sin (P-Q)$$

and also

$$\sin R = \sin (180^\circ-(P+Q))=\color{yellow}\bbox[5px,green]{\sin (P+Q)}$$

Therefore we have $$\sin^2 P -\sin^2 Q=\sin R \sin (P-Q)$$(**)

Replace (**) into (*) we get:

$$\sin R \sin (P-Q)\left(\sin R \sin (P-Q)+\sin^2 Q+\sin P \sin R -\sin^2 Q\right)=\sin^2 Q \sin^2 R$$

$$\sin^2 R \sin (P-Q)\left( \sin (P-Q)+\sin P \right)=\sin^2 Q \sin^2 R$$

$$ \sin (P-Q)\left( \sin (P-Q)+\sin P \right)=\sin^2 Q $$

From the given condition that says $2\angle P=3\angle Q$, we can write $P=3x,\,Q=2x$ for some $x$ so that we are left to prove

$$\sin x(\sin x+\sin 3x)=\sin^2 2x$$

which is true since both sides are equal to $2\sin x \cos x \sin 2x$, this completes the proof.