সৃজনশীলঃ ‘f(x)=sin⁡x,g(x)=cos⁡(x)এবং r=4sin⁡α2sin⁡β2sin⁡γ2−1.…’ ২০২৬ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=\sin x$

$\therefore f\left(y\right)=\sin y$

এবং $g\left(x\right)=\cos x$

$\therefore g\left(y\right)=\cos y$

এখানে, $f\left(x\right)+f\left(y\right)=p$
বা, $\sin x+\sin y=p$
বা, $2\sin \frac{x+y}{2}\cos \frac{x-y}{2}=p$$.....\left(i\right)$

এবং $g\left(x\right)+g\left(y\right)=q$
বা, $\cos x+\cos y=q$
বা, $2\cos \frac{x+y}{2}\cos \frac{x-y}{2}=q$$.....\left(ii\right)$
(i) ও (ii) কে বর্গ করে যোগ করি

$\therefore 4{\cos }^2\frac{x-y}{2}({\sin }^2\frac{x+y}{2}+{\cos }^2\frac{x+y}{2})=p^2+q^2$

বা, $4(1-{\sin }^2\frac{x-y}{2})=p^2+q^2$
বা, $4{\sin }^2\frac{x-y}{2}=4-p^2-q^2$
$\therefore \sin \frac{x-y}{2}=\pm \frac{1}{2}\sqrt{4-p^2-q^2}$. (দেখানো হলো) 

দেওয়া আছে, $\alpha +\beta +\gamma =0$
বা, $\alpha +\beta =-\gamma$
এবং $r=4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}-1$
বামপক্ষ $=\cos \alpha +\cos \beta -\cos \gamma$ 

$\begin{array}{cc}& =\cos \alpha +\cos \beta -\cos (-(\alpha +\beta \left)\right)\\ & =\cos \alpha +\cos \beta -\cos (\alpha +\beta )\\ & =1-2{\sin }^2\frac{\alpha }{2}+1-2{\sin }^2\frac{\beta }{2}-\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ & =2-2{\sin }^2\frac{\alpha }{2}-2{\sin }^2\frac{\beta }{2}-(1-2{\sin }^2\frac{\alpha }{2})(1-2{\sin }^2\frac{\beta }{2})\\ & +4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}\\ & =2-2{\sin }^2\frac{\alpha }{2}-2{\sin }^2\frac{\beta }{2}-1+2{\sin }^2\frac{\alpha }{2}+2{\sin }^2\frac{\beta }{2}\\ & -4{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\beta }{2}+4\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\beta }{2}\\ & =1+4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}(\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2})\\ & =1+4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\cos \frac{\alpha +\beta }{2}\\ & =1+4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\cos (\frac{\pi }{2}-\frac{\gamma }{2})[\because \alpha +\beta +\gamma =0]\\ & =1+4\sin \frac{\alpha }{2}\mathrm{son}\frac{\beta }{2}\sin \frac{\gamma }{2}\\ & =1+4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}-2+2\end{array}$ 

$=4\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\sin \frac{\gamma }{2}-1+2=r+2=$ ডানপক্ষ 

$\therefore \cos \alpha +\cos \beta -\cos \gamma =r+2$. (প্রমাণিত)