সৃজনশীলঃ ‘f(θ)=sin⁡θ;c হলো বৃত্তটির…’ ২০২৫ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $f\left(\theta \right)=\sin \theta$
এখন, $\frac{f(\frac{\pi }{2}-2x)}{f\left(2x\right)}=f\left(x\right)+f(\frac{\pi }{2}+x)$
বা, $\frac{\sin (\frac{\pi }{2}-.2x)}{\sin 2x}=\sin x+\sin (\frac{\pi }{2}+x)$
বা, $\frac{\cos 2x}{\sin 2x}=\sin x+\cos x$
বা, $\cos 2x=\sin 2x(\cos x+\sin x)$
বা, $({\cos }^{2}x-{\sin }^{2}x)-2\sin x\cos x(\cos x+\sin x)=0$
বা, $(\cos x+\sin x)(\cos x-\sin x-2\sin x\cos x)=0$
হয়, $\cos x+\sin x=0$
বা, $1+\tan x=0[\cos x$ দ্বারা ভাগ করে]

বা, $\tan x=-1$ বা, $\tan x=\tan (-\frac{\pi }{4})$$\therefore x=n\pi -\frac{\pi }{4};$ যেখানে $n\in Z$

অথবা, $\cos x-\sin x-2\sin x\cos x=0$

বা, $\cos x-\sin x=2\sin x\cos x$

বা, $\cos x-\sin x=\sin 2x$

বা, $(\cos x-\sin x{)}^2={\sin }^{2}2x$ [বর্গ করে]

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

বা, $1-\sin 2x={\sin }^{2}2x$ বा, ${\sin }^{2}2x+\sin 2x-1=0$

বা, $\sin 2x=\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot (-1)}}{2}=\frac{1}{2}(-1\pm \sqrt{5})$

অর্থাৎ, $\sin 2x=\frac{1}{2}(-1+\sqrt{5})$; কারণ সাইন অনুপাতের মান -1

অপেক্ষা ক্ষুদ্রতর হতে পারে না।

ধরি, $\sin \alpha =\frac{1}{2}(-1+\sqrt{5})$

বা, $\alpha ={\sin }^{-1}\left\{\frac{1}{2}\right(-1+\sqrt{5}\left)\right\}$

তাহলে , $\sin 2x=\sin \alpha$

$\therefore 2x=n\pi +(-1{)}^n\alpha$; যেখানে $n\in Z$

বা, $x=\frac{1}{2}n\pi +(-1{)}^n\cdot \frac{1}{2}\alpha$.

নির্ণেয় সমাধান: $x=n\pi -\frac{\pi }{4},\frac{1}{2}n\pi +(-1{)}^n,\frac{1}{2}\alpha$

যেখানে $\alpha ={\sin }^{-1}\left\{\frac{1}{2}\right(-1+\sqrt{5}\left)\right\}$ 

$C$ কেন্দ্রবিশিষ্ট বৃত্তে, $\beta =2\alpha$ lবৃख্खের একই চাপের উপর দড্ডায়মান কেন্দ্রস্থ কোণ বৃত্তস্থ কোণের দ্মিগুণ]
বা. $\alpha =\frac{\beta }{2}$
দেওয়া আছে, $f\left(\theta \right)=\sin \theta$
বামপক্ষ  $=f(2{\tan }^{-1}(\sqrt{\frac{a}{b}}\tan \alpha ))$ 

$\begin{array}{cc}& =\sin (2{\tan }^{-1}(\frac{\sqrt{a}}{\sqrt{b}}\tan \frac{\beta }{2}))[\because \alpha =\frac{\beta }{2}]\\ & =\sin ({\sin }^{-1}\frac{2\cdot \frac{\sqrt{a}}{\sqrt{b}}\tan \frac{\beta }{2}}{1+{(\frac{\sqrt{a}}{\sqrt{b}}\tan \frac{\beta }{2})}^2})[\because 2{\tan }^{-1}x=\frac{2x}{1+x^2}]\\ & =\frac{\frac{2\sqrt{a}}{\sqrt{b}}\cdot \frac{\sin \frac{\beta }{2}}{\cos \frac{\beta }{2}}}{1+\frac{a}{b}{\tan }^2\frac{\beta }{2}}=\frac{2\sqrt{a}\sin \frac{\beta }{2}}{1+\frac{a}{b}\cdot \frac{\sin \frac{\beta }{2}}{{\mathrm{sos}}^2\frac{\beta }{2}}}=\frac{\frac{2\sqrt{a}\sin \frac{\beta }{2}}{b{\cos }^2\frac{\beta }{2}+a{\sin }^2\frac{\beta }{2}}}{\frac{\sqrt{b}\cos \frac{\beta }{2}}{b{\cos }^2\frac{\beta }{2}}}\end{array}$

$\begin{array}{cc}& =\frac{2\sqrt{a}\sin \frac{\beta }{2}}{\sqrt{b}\cos \frac{\beta }{2}}\times \frac{b{\cos }^2\frac{\beta }{2}}{b{\cos }^2\frac{\beta }{2}+a{\sin }^2\frac{\beta }{2}}\\ & =\frac{2\sqrt{a}\sqrt{b}\sin \frac{\beta }{2}\cos \frac{\beta }{2}}{b{\cos }^2\frac{\beta }{2}+a{\sin }^2\frac{\beta }{2}}=\frac{2(\sqrt{a}\sqrt{b}2\sin \frac{\beta }{2}\cdot \cos \frac{\beta }{2})}{2(b{\cos }^2\frac{\beta }{2}+a{\sin }^2\frac{\beta }{2})}\\ & =\frac{2\sqrt{a}\sqrt{b}\sin (2\cdot \frac{\beta }{2})}{(b\times 2{\cos }^2\frac{\beta }{2})+(a\times 2{\sin }^2\frac{\beta }{2})}[\because 2\sin A\cos A=\sin 2A]\\ & =\frac{2\sqrt{ab}\sin \beta }{b(1+\cos 2\cdot \frac{\beta }{2})+a(1-\cos 2\cdot \frac{\beta }{2})}\end{array}$ 

$[\because 2{\cos }^{2}A=1+\cos 2A$ এবং $2{\sin }^{2}A=1-\cos 2A]$ 

$\begin{array}{cc}& =\frac{2\sqrt{ab}\sin \beta }{b+b\cos \beta +a-a\cos \beta }\\ & =\frac{2\sqrt{ab}\sin \beta }{(b+a)+(b-a)\cos \beta }\end{array}$ 

ডানপক্ষ $=\frac{2\sqrt{ab}f\left(\beta \right)}{(b+a)+(b-a)f(\frac{\pi }{2}-\beta )}$ 

$\begin{array}{cc}& =\frac{2\sqrt{ab}\sin \beta }{(b+a)+(b-a)\sin (\frac{\pi }{2}-\beta )}\\ & =\frac{2\sqrt{ab}\sin \beta }{(b+a)+(b-a)\cos \beta }\end{array}$ 

অর্থাৎ, $f(2{\tan }^{-1}\left(\sqrt{\frac{a}{b}\tan \alpha }\right))=\frac{2\sqrt{ab}f\left(\beta \right)}{(b+a)+(b-a)f(\frac{\pi }{2}-\beta )}$.