# Trigonometric Problem

• February 6th 2011, 11:18 AM
cromlix
Trigonometric Problem
Hi there,
Can anyone give me help to start off this problem please?

Triangle PQR Angle PRQ = 3x; Angle PQR = x

I have to prove that p = 4q cos x cos 2x.

Thanks
Cromlix
• February 6th 2011, 11:29 AM
skeeter
Quote:

Originally Posted by cromlix
Hi there,
Can anyone give me help to start off this problem please?

Triangle PQR Angle PRQ = 3x; Angle PQR = x

I have to prove that p = 4q cos x cos 2x.

Thanks
Cromlix

law of sines ...

$\dfrac{q}{\sin{x}} = \dfrac{p}{\sin(180-4x)}$

$q\sin(180-4x) = p\sin{x}$

... taking it from this point will lead to what you want proved