If $R$ is the circumradius of triangle $ABC$, can we show $a\:/\sin A = 2R$?

If $R$ is the circumradius of triangle $ABC$, can we show $a\:/\sin A = 2R$?

Question

With the usual notation prove that

$$\begin{equation*} \frac{a}{\sin A} = 2R \end{equation*}$$

where $R$ is the radius of the circumcircle of the triangle $ABC$.

[You need consider only the case when the angle $A$ is acute.]

Hence or otherwise deduce that

1. $a \cos A + b \cos B = c \cos (A-B)$,

---

1. $a \cos A + b \cos B + c \cos C = 4 R \sin A \sin B \sin C$.

Solution

In the diagram on the left we have used the facts that

• the perpendicular from the centre bisects the chord $BC$, and

• the angle at the centre is twice the angle at the circumference.

Now we can see that $R \sin A = \dfrac{a}{2}$. Alternatively, in the diagram on the right we have used the facts that

• the angles subtended at the circumference are equal, and

• the angle in a semi-circle is a right-angle.

This shows us that $\sin A = \dfrac{a}{2R}$.

Either of these gives us $\dfrac{a}{\sin A} = 2R$ as required.

We will use the identities $\sin 2x = 2\sin x \cos x$, $\sin (\pi - x) = \sin x$, and $\sin x + \sin y = 2\sin \dfrac{x+y}{2}\cos \dfrac{x-y}{2}$.

Now

$$\begin{align*} a\cos A + b\cos B &= \frac{a}{\sin A} \sin A \cos A + \frac{b}{\sin B} \sin B \cos B\\ &= 2R\sin A \cos A + 2R \sin B \cos B\\ &= R(\sin 2A + \sin 2B)\\ &= R(2\sin (A+B)\cos (A-B))\\ &= 2R\sin (\pi-A-B)\cos (A-B))\\ &= 2R \sin C \cos (A-B)\\ &= \dfrac{c}{\sin C}\sin C \cos (A-B)\\ &= c\cos (A-B), \end{align*}$$

as required.

We’ll use the additional identities $\cos x + \cos y = 2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$, and $\cos\left(\dfrac{\pi}{2}-x\right) = \sin x$.

We have

$$\begin{align*} a \cos A + b \cos B + c \cos C &= \dfrac{a}{\sin A}\sin A \cos A + \dfrac{b}{\sin B}\sin B \cos B + \dfrac{c}{\sin C}\sin C \cos C\\ &= 2R(\sin A \cos A + \sin B \cos B + \sin C \cos C)\\ &= R(\sin 2A + \sin 2B + 2\sin C\cos C)\\ &= R (2\sin (A+B)\cos(A-B) + 2 \sin C \cos C)\\ &= R (2\sin C \cos(A-B) + 2\sin C \cos C)\\ &= 2R \sin C (\cos(A-B)+\cos C)\\ &= 4R \sin C \left(\cos \dfrac{A+C-B}{2} \cos \dfrac{A-B-C}{2}\right)\\ &= 4R \sin C \cos \left(\dfrac{\pi}{2}-B\right) \cos \left(\dfrac{\pi}{2}-A\right)\\ &= 4R \sin A \sin B \sin C. \end{align*}$$