সৃজনশীলঃ ‘দৃশ্যকল্প-১ : M=sin3⁡x+sin3⁡(120∘+x)+sin3⁡(240∘+x)দৃশ্যকল্প-২:…’ ভিকারুননিসা নূন স্কুল এন্ড কলেজ ২০২৬ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $M={\sin }^{3}x+{\sin }^3({120}^{\circ }+x)+{\sin }^3({240}^{\circ }+x)$

বামপক্ষ $={\sin }^{3}x+{\sin }^3({120}^{\circ }+x)+{\sin }^3({240}^{\circ }+x)$

$\begin{array}{cc}=& {\sin }^{3}x+{\sin }^3({180}^{\circ }-{60}^{\circ }+x)+{\sin }^3({180}^{\circ }+{60}^{\circ }+x)\\ =& {\sin }^{3}x+{\sin }^3\{{180}^{\circ }-({60}^{\circ }-x)\}+{\sin }^3\{{180}^{\circ }+({60}^{\circ }+x)\}\\ =& {\sin }^{3}x+{\sin }^3({60}^{\circ }-x)-{\sin }^3({60}^{\circ }+x)\\ =& \frac{1}{4}[4{\sin }^{3}x+4{\sin }^3(60-x)-4{\sin }^3({60}^{\circ }+x)]\\ =& \frac{1}{4}[3\sin x-\sin 3x+3\sin ({60}^{\circ }-x)\\ & -\sin ({180}^{\circ }-3x)-3\sin ({60}^{\circ }+x)+\sin ({180}^{\circ }+3x)]\\ =& \frac{1}{4}[3\sin x-\sin 3x-3\{\sin ({60}^{\circ }+x)-\sin ({60}^{\circ }-x)\}\\ & +\{\sin ({180}^{\circ }+3x)-\sin ({180}^{\circ }-3x)\}]\\ =& \frac{1}{4}[3\sin x-\sin 3x-3.2\cos {60}^{\circ }\sin x+2\cos {180}^{\circ }\sin 3x]\\ =& \frac{1}{4}[3\sin x-\sin 3x-3\sin x+2(-1)\sin 3x]\\ =& \frac{1}{4}[-3\sin 3x]\end{array}$

$=\frac{-3}{4}\sin 3x=$ ডানপক্ষ

$\therefore M=\frac{-3}{4}\sin 3x$. (দেখানো হলো)

দেওয়া আছে, $\frac{\tan \frac{\theta }{2}}{\sqrt{a+b}}=\frac{\tan \frac{\phi }{2}}{\sqrt{a-b}}$

বা, $\frac{{\tan }^2\frac{\theta }{2}}{a+b}=\frac{{\tan }^2\frac{\phi }{2}}{a-b}$

বা, ${\tan }^2\frac{\theta }{2}=\frac{a+b}{a-b}{\tan }^2\frac{\phi }{2}$

বা, $\frac{{\sin }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}=\frac{a+b}{a-b}\frac{{\sin }^2\frac{\phi }{2}}{{\cos }^2\frac{\phi }{2}}$

বা, $\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}=\frac{a+b}{a-b}\frac{2{\sin }^2\frac{\phi }{2}}{2{\cos }^2\frac{\phi }{2}}$

বা, $\frac{1-\cos \theta }{1+\cos \theta }=\frac{a+b}{a-b}\frac{1-\cos \phi }{1+\cos \phi }$

বা, $\frac{1+\cos \theta }{1-\cos \theta }=\frac{a-b}{a+b}\frac{1+\cos \phi }{1-\cos \phi }$

বা, $\frac{1+\cos \theta }{\cos \theta -1}=\frac{a-b}{-(a+b)}\frac{1+\cos \phi }{1-\cos \phi }$

বা, $\frac{1+\cos \theta +\cos \theta -1}{1+\cos \theta -\cos \theta +1}=\frac{(a-b)(1+\cos \phi )-(a+b)(1-\cos \phi )}{(a-b)(1+\cos \phi )+(a+b)(1-\cos \phi )}$

[যোজন-বিয়োজন করে]

বা, $\frac{2\cos \theta }{2}=\frac{a-b+(a-b)\cos \phi -a-b+(a+b)\cos \phi }{a-b+(a-b)\cos \phi +a+b-(a+b)\cos \phi }$

বা, $\cos \theta =\frac{-2b+2a\cos \phi }{2a-2b\cos \phi }$

বা, $\cos \theta =\frac{2(a\cos \phi -b)}{2(a-b\cos \phi )}$

$\therefore \cos \theta =\frac{a\cos \phi -b}{a-b\cos \phi }$ (প্রমাণিত)