সৃজনশীলঃ ‘p=sin⁡2α,q=sin⁡2β,r=cos⁡2α,s=cos⁡2β,t=sin⁡2γ.…’ Combined ২০১৮ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $p=\sin 2\alpha ,q=\sin 2\beta ,r=\cos 2\alpha ,s=\cos 2\beta$

$p+q=c$

$\therefore \sin 2\alpha +\sin 2\beta =c..........\left(1\right)$

এবং $r+s=d$

$\therefore \cos 2\alpha +\cos 2\beta =d........\left(2\right)$

(2) নং কে (1) নং দ্বারা ভাগ করে পাই,

$\frac{\cos 2\alpha +\cos 2\beta }{\sin 2\alpha +\sin 2\beta }=\frac{d}{c}$

বা, $\frac{2\cos \frac{2\alpha +2\beta }{2}\cos \frac{2\alpha -2\beta }{2}}{2\sin \frac{2\alpha +2\beta }{2}\cos \frac{2\alpha -2\beta }{2}}=\frac{d}{c}$

বা, $\frac{\cos (\alpha +\beta )}{\sin (\alpha +\beta )}=\frac{d}{c}$

বা, $\frac{{\cos }^2(\alpha +\beta )}{{\sin }^2(\alpha +\beta )}=\frac{d^2}{e^2}$ [বর্গ করি]

বা, $\frac{{\cos }^2(\alpha +\beta )+{\sin }^2(\alpha +\beta )}{{\cos }^2(\alpha +\beta )-{\sin }^2(\alpha +\beta )}=\frac{d^2+c^2}{d^2-c^2}$ [যোজন বিয়োজন করে]

বা, $\frac{1}{\cos 2(\alpha +\beta )}=\frac{d^2+c^2}{d^2-c^2}$

বা, $\frac{1}{\cos (2\alpha +2\beta )}=\frac{d^2+c^2}{d^2-c^2}$

$\therefore \cos (2\alpha +2\beta )=\frac{d^2-c^2}{d^2+c^2}$. (দেখানো হলো)

দেওয়া আছে, $p=\sin 2\alpha$

$q=\sin 2\beta$

$t=\sin 2\gamma$

এবং $\alpha +\beta +\gamma =\pi$

বামপক $=p^2+q^2+t^2$

$={\sin }^{2}2\alpha +{\sin }^{2}2\beta +{\sin }^{2}2\gamma$

$=\frac{1}{2}(2{\sin }^{2}2\alpha +2{\sin }^{2}2\beta )+{\sin }^{2}2\gamma$

$=\frac{1}{2}(1-\cos 4\alpha +1-\cos 4\beta )+{\sin }^{2}2\gamma$

$=1-\frac{1}{2}(\cos 4\alpha +\cos 4\beta )+{\sin }^{2}2\gamma$

$=1-\frac{1}{2}\cdot 2\cos \frac{4\alpha +4\beta }{2}\cos \frac{4\alpha -4\beta }{2}+{\sin }^{2}2\gamma$

$=1-\cos (2\alpha +2\beta )\cos (2\alpha -2\beta )+{\sin }^{2}2\gamma$

$=1-\cos (2\pi -2\gamma )\cos (2\alpha -2\beta )+1-{\cos }^{2}2\gamma$

$[\because \alpha +\beta +\gamma =\pi$

$\therefore 2\alpha +2\beta =2\pi -2\gamma ]$

$=2-\cos 2\gamma \cos (2\alpha -2\beta )-\cos 2\gamma$

$=2-\cos 2\gamma \{\cos (2\alpha -2\beta )+\cos 2\gamma \}$

$=2-\cos 2\gamma \{\cos (2\alpha -2\beta )+\cos (2\pi -2\alpha -2\beta \left)\right\}$

$=2-\cos 2\gamma \{\cos (2\alpha -2\beta )+\cos (2\alpha +2\beta \left)\right\}$

$=2-\cos 2\gamma \cdot 2\cos 2\alpha \cos 2\beta$

$=2-2\cos 2\alpha \cos 2\beta \cos 2\gamma =$ ডানপক্ষ

$\therefore p^2+q^2+t^2=2-2\cos 2\alpha \cos 2\beta \cos 2\gamma .$ (দেখানো হলো)