সৃজনশীলঃ ‘দৃশ্যকল্প-১ : f(x)=sin⁡x.…’ ২০২৬ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$\begin{array}{cc}& A+B+C=\pi ,f\left(x\right)=\sin x\\ \therefore & f\left(4A\right)=\sin 4A,f\left(4B\right)=\sin 4B,f\left(4C\right)=\sin 4C\\ & f\left(2A\right)=\sin 2A,f\left(2B\right)=\sin 2B,f\left(2C\right)=\sin 2C\end{array}$ 

বামপক্ষ $=f\left(4A\right)+f\left(4B\right)+f\left(4C\right)$ 

$\begin{array}{cc}& =\sin 4A+\sin 4B+\sin 4C\\ & =2\sin (2A+2B)\cos (2A-2B)+\sin 4C\\ & =2\sin (2\pi -2C)\cos (2A-2B)+\sin 4C\\ & =-2\sin 2C\cos (2A-2B)+\sin 2\cdot 2C\\ & =-2\sin 2C\cos (2A-2B)+2\sin 2C\cos 2C\\ & =-2\sin 2C[\cos (2A-2B)-\cos 2C]\\ & =-2\sin 2C[\cos (2A-2B)-\cos \{2\pi -(2A+2B)\left\}\right]\\ & =-2\sin 2C[\cos (2A-2B)-\cos (2A+2B\left)\right]\\ & =-2\sin 2C\cdot 2\sin 2A\sin 2B\\ & =-4\sin 2A\sin 2B\sin 2C\end{array}$ 

$=-4f\left(2A\right)f\left(2B\right)f\left(2C\right)=$ ডানপক্ষ 

দেওয়া আছে, $A={15}^{\circ }$
বামপক্ষ $=b^4+c^4+a^4-\sqrt{3}{b}^2{c}^2$ 

$\begin{array}{cc}& ={\left(b^2\right)}^2+{\left(c^2\right)}^2+{(-a^2)}^2+2b^2{c}^2-2a^2{b}^2-2c^2{a}^2\\ & ={(b^2+c^2-a^2)}^2-\sqrt{3}{b}^2{c}^2-2b^2-2b^2{c}^2+2a^2{a}^2(b^2+c^2)\\ & ={(\frac{b^2+c^2-a^2}{2bc}\times 2bc)}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\\ & =(\cos A\times 2bc{)}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\\ & ={(\cos {15}^{\circ }\times 2bc)}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\\ & ={\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^2\times 4b^2{c}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\\ & =\frac{3+1+2\sqrt{3}}{4\times 2}\times 4b^2{c}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\\ & =(2+\sqrt{3})b^2{c}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\\ & =2b^2{c}^2+\sqrt{3}{b}^2{c}^2-\sqrt{3}{b}^2{c}^2-2b^2{c}^2+2a^2(b^2+c^2)\end{array}$ 

$=2a^2(b^2+c^2)=$ ডানপক্ষ 

$\therefore b^4+c^4+a^4-\sqrt{3}{b}^2{c}^2=2a^2(b^2+c^2)$. (প্রমাণিত)