Question

If $\mathrm{P},\mathrm{Q}$ and $\mathrm{R}$ are the interior angles of $∆\mathrm{PQR}$, then show that :
$\cos \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)\sin \left(\frac{\mathrm{P}}{2}\right)+\sin \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)\cos \left(\frac{\mathrm{P}}{2}\right)=1$

Answer 1

Using the angle sum property of a triangle and the trigonometric identities:

$\angle P+\angle Q+\angle R=180°---\left(1\right)$

Dividing $\left(1\right)$ by 2,

$\frac{P+Q+R}{2}=\frac{180°}{2}=90°$

$\frac{Q+R}{2}=90°-\frac{P}{2}...\left(4\right)$

Now,

$\cos \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)\sin \left(\frac{\mathrm{P}}{2}\right)+\sin \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)\cos \left(\frac{\mathrm{P}}{2}\right)=1$

$\cos \left(90°-\frac{P}{2}\right)\sin \left(\frac{\mathrm{P}}{2}\right)+\sin \left(90°-\frac{P}{2}\right)\cos \left(\frac{\mathrm{P}}{2}\right)=1\left[\text{using equation}\left(4\right)\right]$

$\sin \left(\frac{P}{2}\right)\sin \left(\frac{\mathrm{P}}{2}\right)+\cos \left(\frac{P}{2}\right)\cos \left(\frac{\mathrm{P}}{2}\right)=1\left[\because \cos (90-A)=\sin A,\sin (90-A)=\cos A\right]$

${\sin }^2\left(\frac{\mathrm{P}}{2}\right)+{\cos }^2\left(\frac{\mathrm{P}}{2}\right)=1=RHS\left[\because {\sin }^{2}A+{\cos }^{2}A=1\right]$

Since, $LHS=RHS$.

Hence, the given expression $\cos \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)\sin \left(\frac{\mathrm{P}}{2}\right)+\sin \left(\frac{\mathrm{Q}+\mathrm{R}}{2}\right)\cos \left(\frac{\mathrm{P}}{2}\right)=1$ proved.