Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Question

Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Can I get a hint so that I can get started? I tried using the standard half angle formula, but they don’t work.

Answer 1

Using the Half Angle Formula,

$$\cos^2 \frac{R}{2} = \dfrac{s(s -r)}{pq}$$

where $s = \dfrac{p+q+r}{2}$

$$\cos^2 \frac{P}{2} = \dfrac{s(s -p)}{rq}$$

$$p\cos^2 \frac{R}{2}+r\cos^2 \frac{P}{2} =\dfrac{s(s -r)}{q} +\dfrac{s(s -p)}{q}=\dfrac{s(2s-(r+p))}{q} = \dfrac{s(q)}{q}=s=\dfrac{p+q+r}2{}$$

Answer 2

By the cosine rule,$$\cos^2\frac{R}{2}=\frac{1+\cos R}{2}=\frac{p^2+q^2-r^2+2pq}{4pq}$$so$$p\cos^2\frac{R}{2}=\frac{p^3r+pq^2r-pr^3+2p^2qr}{4pqr}.$$Add $p\leftrightarrow r$, then simplify.

Answer 3

\begin{align} \dfrac{p+q+r}{2} &= p\cos^2\frac R2+r\cos^2\frac P2\\ &= \dfrac{p+p\cos R+r\cos P+r}{2} \end{align} which happens iff $$q = p\cos R+r\cos P$$ This is evident from the geometry in the following image:

Answer 4

The standard half-angle formula $\cos^2 \frac x2 = \frac{1+\cos x}2$ is helpful if combined with the law of cosines.

Using above half-angle formula, the equation to be shown is equivalent with

$$p\cos R+ r \cos P = q$$

Now, just replace $\cos R = \frac{q^2+p^2-r^2}{2pq}$ and $\cos P = \frac{q^2+r^2-p^2}{2qr}$: \begin{eqnarray*} p\cos R+ r \cos P & = & \frac 1{2q}\left(q^2+p^2-r^2+q^2+r^2-p^2\right)\\ & = & q \end{eqnarray*}