সৃজনশীলঃ ‘উদ্দীপক-১: h(x)=cos⁡x.উদ্দীপক-২: এ△ABC−এa=3+1,…’ নটরডেম কলেজ ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$\phi \left(x\right)=\cos x$

$\phi \left(4x\right)=\cos 4x$

$\phi \left(8x\right)=\cos 8x$

$\phi \left(14x\right)=\cos 14x$

প্রদত্ত রাশি, $=\phi \left(2x\right)\cdot \phi \left(4x\right)\cdot \phi \left(8x\right)\cdot \phi \left(14x\right)$

$=\cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{2\sin 2x}(2\sin 2x\cdot \cos 2x)\cdot \cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{2\sin 2x}\sin 4x\cdot \cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{4\sin 2x}\cdot 2\sin 4x\cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{4\sin 2x}\cdot \sin 8x\cdot \cos 8x\cdot \cos 14x$

$\begin{array}{cc}& =\frac{1}{8\sin 2x}\cdot 2\sin 8x\cdot \cos 8x\cdot \cos 14x\\ & =\frac{1}{8\sin 2x}\cdot \sin 16x\cdot \cos 14x\\ & =\frac{1}{16\sin 2x}\cdot 2\sin 16x\cdot \cos 14x\\ & =\frac{1}{16\sin 2x}\{\sin (16x+14x)+\sin (16x-14x\left)\right\}\\ & =\frac{1}{16\sin 2x}(\sin 30x+\sin 2x)\\ & =\frac{1}{16\sin \frac{2\pi }{15}}(\sin 2\pi +\sin \frac{2\pi }{15})[\because x=\frac{\pi }{15}]\\ & =\frac{1}{16\sin \frac{2\pi }{15}}(0+\sin \frac{2\pi }{15})=\frac{1}{16}\end{array}$

নির্ণেয় মান : $\frac{1}{16}$

দেওয়া আছে, $△\mathrm{ABC}-△a=\sqrt{3}+1$

$b=\sqrt{3}-1$ এবং $C={60}^{\circ }$

আমরা জানি, $\cos C=\frac{a^2+b^2-c^2}{2ab}$

বা, $\cos {60}^{\circ }=\frac{(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-c^2}{2(\sqrt{3}+1)\cdot (\sqrt{3}-1)}$

বা, $\frac{1}{2}=\frac{2\cdot 3+2\cdot 1-c^2}{2(3-1)}[\because (a+b{)}^2+(a-b{)}^2=2a^2+2b^2]$

বা, $1=\frac{8-c^2}{2}$

বা, $2=8-c^2$

বা, $c^2=6$

$\therefore c=\sqrt{6}$ একক

$△\mathrm{ABC}$ এর অর্ধ-পরিসীমা,

$S=\frac{a+b+c}{2}=\frac{\sqrt{3}+1+\sqrt{3}-1+\sqrt{6}}{2}$

$=\frac{2\sqrt{3}+\sqrt{6}}{2}=\sqrt{3}+\frac{\sqrt{6}}{2}$

$\Delta \mathrm{ABC}$ এর ক্ষেত্রফল, $\Delta =\sqrt{S(S-a)(S-b)(S-c)}$

$=\sqrt{(\sqrt{3}+\frac{\sqrt{6}}{2})(\sqrt{3}+\frac{\sqrt{6}}{2}-\sqrt{3}-1)(\sqrt{3}+\frac{\sqrt{6}}{2}-\sqrt{3}+1)(\sqrt{3}+\frac{\sqrt{6}}{2}-\sqrt{6})}$

$=\sqrt{(\sqrt{3}+\frac{\sqrt{6}}{2})(\frac{\sqrt{6}}{2}-1)(\frac{\sqrt{6}}{2}+1)(\sqrt{3}-\frac{\sqrt{6}}{2})}$

$=\sqrt{\left\{\right(\sqrt{3}{)}^2-{\left(\frac{\sqrt{6}}{2}\right)}^2\}\{{\left(\frac{\sqrt{6}}{2}\right)}^2-1^2\}}$

$=\sqrt{(3-\frac{6}{4})(\frac{6}{4}-1)}=\sqrt{\frac{6}{4}\times \frac{2}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

$\therefore \mathrm{ABC}$ এর অন্তঃব্যাসার্ধ

$r=\frac{\Delta }{S}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{3}+\frac{\sqrt{6}}{2}}=\frac{\sqrt{3}}{2\sqrt{3}+\sqrt{6}}$

$=\frac{\sqrt{3}}{\sqrt{3}(2+\sqrt{2})}=\frac{1}{2+\sqrt{2}}$

$=\frac{2-\sqrt{2}}{(2+\sqrt{2})(2-\sqrt{2})}$

$=\frac{2-\sqrt{2}}{2^2-(\sqrt{2}{)}^2}=\frac{2-\sqrt{2}}{4-2}=\frac{2-\sqrt{2}}{2}$

নির্ণেয় অন্তঃব্যাসার্ধ $\frac{2-\sqrt{2}}{2}$ একক।