সৃজনশীলঃ ‘দৃশ্যকল্প-১: f(θ)=sin⁡θ.দৃশ্যকল্প-২: A=cosec−15−12sin−1⁡35+tan−1⁡14…’ যশোর বোর্ড ২০২১ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(\theta \right)=\sin \theta$

প্রদত্ত সমীকরণ,

$2f(\frac{\pi }{2}-\theta )\cdot f(\frac{\pi }{2}-3\theta )+1=0$

$\Rightarrow 2\cos \theta \cdot \cos 3\theta +1=0$

$\Rightarrow \cos (3\theta +\theta )+\cos (3\theta -\theta )+1=0$

$\begin{array}{cc}& \Rightarrow \cos 4\theta +\cos 2\theta +1=0\\ & \Rightarrow 2{\cos }^{2}2\theta +\cos 2\theta =0\\ & \Rightarrow \cos 2\theta (2\cos 2\theta +1)=0\\ & \therefore \cos 2\theta =0\\ & \Rightarrow 2\theta =(2n+1)\frac{\pi }{2}\\ & \Rightarrow \theta =(2n+1)\frac{\pi }{4},nϵZ\end{array}$

অথবা, $\cos 2\theta =-\frac{1}{2}=\cos \frac{2\pi }{3}$

$\begin{array}{cc}& \Rightarrow 2\theta =2n\pi \pm \frac{2\pi }{3}\\ & \therefore \theta =n\pi \pm \frac{\pi }{3},n\in Z\end{array}$

$\therefore$ নির্ণেয় সমাধানঃ $x=(2n+1)\frac{\pi }{4},n\pi \pm \frac{\pi }{3},n\in Z$

দৃশ্যকল্প-২ হতে,

$\begin{array}{cc}& A={\mathrm{cosec}}^{-1}\sqrt{5}-\frac{1}{2}{\sin }^{-1}\frac{3}{5}\\ & +{\tan }^{-1}\frac{1}{4}\end{array}$

ধরি $\frac{1}{2}{\sin }^{-1}\frac{3}{5}=\theta$

$\begin{array}{cc}& \Rightarrow {\sin }^{-1}\frac{3}{5}=2\theta \\ & \Rightarrow \sin 2\theta =\frac{3}{5}\end{array}$

$\begin{array}{cc}\therefore \tan \theta & =\frac{\sin \theta }{\cos \theta }\\ & =\frac{2\sin \theta \cdot \sin \theta }{2\sin \theta \cdot \cos \theta }\end{array}$

$\begin{array}{cc}& =\frac{2{\sin }^2\theta }{\sin 2\theta }\\ & =\frac{1-\cos 2\theta }{\sin 2\theta }\\ & =\frac{1-\sqrt{1-{\sin }^{2}2\theta }}{\sin 2\theta }\\ & =\frac{1-\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}\\ & =\frac{1-\frac{4}{5}}{\frac{3}{5}}\\ & =\frac{\frac{1}{5}}{\frac{3}{5}}\\ & =\frac{1}{3}\\ & \therefore \theta ={\tan }^{-1}\frac{1}{3}\\ & \therefore A={\mathrm{cosec}}^{-1}\sqrt{5}-\frac{1}{2}{\sin }^{-1}\frac{3}{5}\\ & +{\tan }^{-1}\frac{1}{4}\\ & ={\tan }^{-1}\frac{1}{2}-{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{4}\\ & ={\tan }^{-1}\frac{\frac{1}{2}-\frac{1}{3}}{1+\frac{1}{2}\cdot \frac{1}{3}}+{\tan }^{-1}\frac{1}{4}\\ & ={\tan }^{-1}\frac{\frac{1}{6}}{\frac{7}{6}}+{\tan }^{-1}\frac{1}{4}\\ & ={\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{4}\\ & ={\tan }^{-1}\frac{\frac{1}{7}+\frac{1}{4}}{1-\frac{1}{7}\cdot \frac{1}{4}}\\ & ={\tan }^{-1}\frac{11}{27}\end{array}$

$=R.H.S$ (দেখানো হল)