# Thread: From Sine Rule to cos (theta) = p / 2q

1. ## From Sine Rule to cos (theta) = p / 2q

Hi,

Doing some self-teach on Maths, and got stumped by the following:-

In triangle PQR, angle PQR = theta and angle QPR = 2 theta. Prove cos (theta) = p / 2q

This question occurs before presentation of trignometric identities, so it implies that knowledge of these is not required to solve the above.

The question is given as part of a discussion of the Sine Rule, given in the form

a / sin A = b / sin B = c / sin C

I've perhaps gone down the wrong the wrong route by:-

• Making P the centre of a circle
• R lie on the circumference of a circle
• Q a point in the circle (and perhaps on)
• Drawing a line of length p from R to interesct the circle, drawing a diameter of length 2q from R to P and onto another point, giving a right angle triangle with a hypotenuse 2q and base p, with an angle whose cosine is p / 2q

What I cannot work out is a way to relate the angle of the triangle I constructed to the angles given in the original triangle.

( The text derives the Sine rule from i) angle in a semicircle, ii) angles in same segment, iii) opposite angles of a cyclic quadrilateral).

Knowing where I'm going wrong with this would be most helpful. Thanks!

2. ## Re: From Sine Rule to cos (theta) = p / 2q

You need not draw any circles! The problem is quite simple.

According to sine rule:

$\frac{q}{\sin{\theta}}=\frac{p}{\sin{(2\theta)}}= \frac{r}{\sin(\hat{R})}$

We have : . $\frac{q}{\sin{\theta}}=\frac{p}{\sin{(2\theta)}}$

Using double angle formula we get :.. $\frac{q}{\sin{\theta}}=\frac{p}{2\sin(\theta)\cos( \theta)}$

$\implies q=\frac{p}{2\cos(\theta)}$

$\implies \cos(\theta)=\frac{p}{2q}$