সৃজনশীলঃ ‘sin⁡x−sin⁡y=α,cos⁡x−cos⁡y=β…’ সরকারি মাইকেল মধুসূদন (এম.এম) কলেজ ২০২৬ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $\sin x-\sin y=\alpha$ এরং $\cos x-\cos y=\beta$

ডানপক্ষ $=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2-{\beta }^2}$
$=\frac{(\sin x-\sin y{)}^2+(\cos x-\cos y{)}^2}{(\sin x-\sin y{)}^2-(\cos x-\cos y{)}^2}$
$=\frac{{\sin }^{2}x-2\sin x\sin y+{\sin }^{2}y+{\cos }^{2}x-2\cos x\cos y+{\cos }^{2}y}{{\sin }^{2}x-2\sin x\sin y+{\sin }^{2}y-{\cos }^{2}x+2\cos x\cos y-{\cos }^{2}y}$
$=\frac{({\sin }^{2}x+{\cos }^{2}x)+({\sin }^{2}y+{\cos }^{2}y)-2(\sin x\sin y+\cos x\cos y)}{({\sin }^{2}x-{\cos }^{2}x)+({\sin }^{2}y-{\cos }^{2}y)-2(\sin x\sin y+\cos x\cos y)}$
$=\frac{2-2\{\sin x\sin y+\cos x\cos y\}}{({\sin }^{2}x-{\cos }^{2}x)+({\sin }^{2}y-{\cos }^{2}y)+2(\cos x\cos y-\sin x\sin y)}$
$=\frac{2-2\cos (x-y)}{-\cos 2x-\cos 2y+2\cos (x+y)}$ 
$=\frac{2\{1-\cos (x-y\left)\right\}}{-2\cos (x+y)\cos (x-y)+2\cos (x+y)}$
$=\frac{2\{1-\cos (x-y\left)\right\}}{2\cos (x+y)\{1-2\cos (x-y\left)\right\}}$
$=\frac{1}{\cos (x+y)}=\sec (x+y)=$ বামপক্ষ 
$\therefore \frac{{\alpha }^2+{\beta }^2}{{\alpha }^2-{\beta }^2}=\sec (x+y)$ (দেখানো হলো)

দেওয়া আছে, $\sin x-\sin y=\alpha ,\cos x-\cos y=\beta$ এবং $\alpha =\beta$ হলে,

$\sin x-\sin y=\cos x-\cos y$

বা, $2\cos \frac{x+y}{2}\sin \frac{x-y}{2}=2\sin \frac{x+y}{2}\sin \frac{y-x}{2}$
বা, $\cos \frac{x+y}{2}\sin \frac{x-y}{2}=-\sin \frac{x+y}{2}\sin \frac{x-y}{2}$

$\therefore \cos \frac{x+y}{2}=-\sin \frac{x+y}{2}$

এখন, ${\alpha }^2=(\sin x-\sin y{)}^2={(2\cos \frac{x+y}{2}\sin \frac{x-y}{2})}^2$

$=4{\cos }^2\frac{x+y}{2}{\sin }^2\frac{x-y}{2}$

$=4{\sin }^2\frac{x+y}{2}{\sin }^2\frac{x-y}{2}[\because \cos \frac{x+y}{2}=\sin \frac{x+y}{2}]$
$=4\times \frac{1-\cos (x-4)}{2}{\sin }^2\frac{x+y}{2}$
$=2\{1-\cos (x-y\left)\right\}{\sin }^2\frac{x+y}{2}$

এখন, ${\sin }^2\frac{x+y}{2}+{\cos }^2\frac{x+y}{2}=1$
বা, ${\sin }^2\frac{x+y}{2}+{\sin }^2\frac{x+y}{2}=1$
বা, $2{\sin }^2\frac{x+y}{2}=1$
বা, ${\sin }^2\frac{x+y}{2}=\frac{1}{2}$

$\therefore {\alpha }^2=2\{1-\cos (x-y\left)\right\}\frac{1}{2}=1-\cos (x-y)$
$\therefore \cos (x-y)=1-{\alpha }^2.$ (প্রমাণিত)