সৃজনশীলঃ ‘f(x)=sin⁡x এবং g(y)=cos⁡y…’ যশোর বোর্ড ২০২২ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

এখানে, $f\left(x\right)=\sin x$ এবং $g\left(y\right)=\cos y$

$\therefore f\left(3x\right)=\sin 3x\therefore g\left(y\right)=\cos x$

এবং $f(\frac{\pi }{2}-2x)=\sin (\frac{\pi }{2}-2x)$ এবং $g(\frac{\pi }{2}-2x)=\cos (\frac{\pi }{2}-2x)$

$=\cos 2x$$$$=\sin 2x$

প্রদত্ত সমীকরণ, $f\left(x\right)+g(\frac{\pi }{2}-2x)+f\left(3x\right)=1+g\left(x\right)+f(\frac{\pi }{2}-2x)$

বা, $\sin x+\sin 2x+\sin 3x=1+\cos x+\cos 2x$ বা, $\sin 3x+\sin x+\sin 2x=1+\cos 2x+\cos x$

বা, $2\cdot \sin \frac{3x+x}{2}\cos \frac{3x-x}{2}+\sin 2x=2{\cos }^{2}x+\cos x$

বা, $2\sin 2x\cos x+\sin 2x=\cos x(2\cos x+1)$

বা, $\sin 2x(2\cos x+1)=\cos x(2\cos x+1)$

বা, $\sin 2x(2\cos x+1)-\cos x(2\cos x+1)$

বা, $(2\cos x+1)(\sin 2x-\cos x)=0$

হয়, $2\cos x+1=0$

অথবা, $\sin 2x-\cos x=0$

বা, $\cos x=-\frac{1}{2}$

বা$\text{ বা, }2\sin x\cos x-\cos x=0$

বা, $\cos x(2\sin x-1)=0$

বা, $\cos x=-\cos \frac{\pi }{3}$ আবার, $\cos x=0$ অথবা, $2\sin x-1=0$

বা, $\cos x=\cos (\pi -\frac{\pi }{3})$ বা, $\cos x=\cos \frac{\pi }{2}$ বা, $\sin x=\frac{1}{2}$

বা, $\cos x=\cos \frac{2\pi }{3}\therefore x=(2n+1)\frac{\pi }{2}$ বা, $\sin x=\sin \frac{\pi }{6}$

$\therefore x=2n\pi \pm \frac{2\pi }{3}$ যেখানে $n$ পূর্ণসংখ্যা $\therefore x=n\pi +(-1{)}^n\frac{\pi }{6}$

যেখানে $n$ পূর্ণসংখ্যা $$যেখানে $n$ পূর্ণসংখ্যা

নির্ণেয় সমাধান, $(2n+1)\frac{\pi }{2},2n\pi \pm \frac{2\pi }{3},n\pi +(-1{)}^n\frac{\pi }{6}$;

যেখানে $n$ যেকোনো পূর্ণসংখ্যা।

এখানে, $f\left(x\right)=\sin x$ এবং $g\left(y\right)=\cos y$

সুতরাং, $2{\tan }^{-1}\frac{f\left(\frac{\alpha }{2}\right)}{f(\frac{\pi }{2}-\frac{\alpha }{2})}\tan (\frac{\pi }{4}-\frac{\beta }{2})={\tan }^{-1}\frac{f\left(\alpha \right)\cdot f\left(\beta \right)}{g(\frac{\pi }{2}-\beta )+f(\frac{\pi }{2}-\alpha )}$

বা, $2{\tan }^{-1}\frac{\sin \frac{\alpha }{2}}{\sin (\frac{\pi }{2}-\frac{\alpha }{2})}\tan (\frac{\pi }{4}-\frac{\beta }{2})={\tan }^{-1}\frac{\sin \alpha \cdot \cos \beta }{\cos (\frac{\pi }{2}-\beta )+\sin (\frac{\pi }{2}-\alpha )}$

বা, $2{\tan }^{-1}\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}}\tan (\frac{\pi }{4}-\frac{\beta }{2})={\tan }^{-1}\frac{\sin \alpha \cdot \cos \beta }{\sin \beta +\cos \alpha }$

বামপক্ষ $=2{\tan }^{-1}\frac{\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}}\tan (\frac{\pi }{4}-\frac{\beta }{2})$

$\begin{array}{cc}& =2{\tan }^{-1}\frac{\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})}{\cos \frac{\alpha }{2}\cdot \cos (\frac{\pi }{4}-\frac{\beta }{2})}={\tan }^{-1}\frac{2\cdot \frac{\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})}{\cos \frac{\alpha }{2}\cdot \cos (\frac{\pi }{4}-\frac{\beta }{2})}}{1-\frac{{\sin }^2\frac{\alpha }{2}{\sin }^2(\frac{\pi }{4}-\frac{\beta }{2})}{{\cos }^2\frac{\alpha }{2}{\cos }^2(\frac{\pi }{4}-\frac{\beta }{2})}}\\ & ={\tan }^{-1}\frac{2\cdot \frac{\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})}{\cos \frac{\alpha }{2}\cdot \cos (\frac{\pi }{4}-\frac{\beta }{2})}}{\frac{{\cos }^2\frac{\alpha }{2}{\cos }^2(\frac{\pi }{4}-\frac{\beta }{2})-{\sin }^2\frac{\alpha }{2}{\sin }^2(\frac{\pi }{4}-\frac{\beta }{2})}{{\cos }^2\frac{\alpha }{2}{\sin }^2(\frac{\pi }{4}-\frac{\beta }{2})}}\\ & ={\tan }^{-1}\frac{2\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})\cdot \cos \frac{\alpha }{2}\cos (\frac{\pi }{4}-\frac{\beta }{2})}{{\{\cos \frac{\alpha }{2}\cos (\frac{\pi }{4}-\frac{\beta }{2})\}}^2-{\{\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})\}}^2}\\ & ={\tan }^{-1}\frac{\frac{1}{2}\cdot 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\cdot 2\sin (\frac{\pi }{4}-\frac{\beta }{2})\cos (\frac{\pi }{4}-\frac{\beta }{2})}{\{\cos \frac{\alpha }{2}\cos (\frac{\pi }{4}-\frac{\beta }{2})+\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})\}\cos \frac{\alpha }{2}\cos (\frac{\pi }{4}-\frac{\beta }{2})-\sin \frac{\alpha }{2}\sin (\frac{\pi }{4}-\frac{\beta }{2})\}}\\ & ={\tan }^{-1}\frac{\sin \alpha \sin 2(\frac{\pi }{4}-\frac{\beta }{2})}{\cos \alpha +\cos (-\frac{\pi }{2}+\beta )}\\ & ={\tan }^{-1}\frac{\sin \alpha \sin (\frac{\pi }{2}-\beta )}{\cos (\frac{\alpha }{2}-\frac{\pi }{4}+\frac{\beta }{2}+\frac{\alpha }{2}+\frac{\pi }{4}-\frac{\beta }{2})+\cos (\frac{\alpha }{2}-\frac{\pi }{4}+\frac{\beta }{2}-\frac{\alpha }{2}-\frac{\pi }{4}+\frac{\beta }{2})}\\ & ={\tan }^{-1}\frac{\sin \alpha \cdot \cos \beta }{\cos \alpha +\cos (-\frac{\pi }{2}+\beta )}\\ & ={\tan }^{-1}\frac{\sin \alpha \cos \beta }{\cos \alpha +\cos (\frac{\pi }{2}-\beta )}[\because \cos (-\theta )=\cos \theta ]\\ & ={\tan }^{-1}\frac{\sin \alpha \cdot \cos \beta }{\sin \beta +\cos \alpha }\end{array}$