সৃজনশীলঃ ‘f(x)=cos⁡x…’ রংপুর ক্যাডেট কলেজ ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$f\left(x\right)=\cos x$

$\therefore f\left(4x\right)=\cos 4x$

$f\left(8x\right)=\cos 8x$

$f\left(14x\right)=\cos 14x$

প্রদত্ত রाশि $=f\left(2x\right)\cdot f\left(4x\right)\cdot f\left(8x\right)\cdot f\left(14x\right)$

$=\cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{2\sin 2x}(2\sin 2x\cdot \cos 2x)\cdot \cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{2\sin 2x}\sin 4x\cdot \cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{4\sin 2x}\cdot 2\sin 4x\cos 4x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{4\sin 2x}\cdot \sin 8x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{8\sin 2x}\cdot 2\sin 8x\cdot \cos 8x\cdot \cos 14x$

$=\frac{1}{8\sin 2x}\cdot \sin 16x\cdot \cos 14x$

$=\frac{1}{16\sin 2x}\cdot 2\sin 16x\cdot \cos 14x$

$=\frac{1}{16\sin 2x}\{\sin (16x+14x)+\sin (16x-14x\left)\right\}$

$=\frac{1}{16\sin 2x}(\sin 30x+\sin 2x)$

$=\frac{1}{16\sin \frac{2\pi }{15}}(\sin 2\pi +\sin \frac{2\pi }{15})[\because x=\frac{\pi }{15}]$

$=\frac{1}{16\sin \frac{2\pi }{15}}(0+\sin \frac{2\pi }{15})=\frac{1}{16}$

নির্ণেয় মান : $\frac{1}{16}$.

দেওয়া আছে, $f\left(x\right)=\cos x$

$\therefore f\left(\alpha \right)=\cos \alpha$

$\therefore f\left(\beta \right)=\cos \beta$

$f(\frac{\pi }{2}-\alpha )=\cos (\frac{\pi }{2}-\alpha )=\sin \alpha$

$\therefore f(\frac{\pi }{2}-\beta )=\cos (\frac{\pi }{2}-\beta )=\sin \beta$

এখানে, $f(\frac{\pi }{2}-\alpha )+f(\frac{\pi }{2}-\beta )=s$

$\therefore \sin \alpha +\sin \beta =s$

$\therefore 2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=s$ ........ (i)

আবার, $f\left(\alpha \right)+f\left(\beta \right)=t$

বা, $\cos \alpha +\cos \beta =t$

$\therefore 2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=t.$

(i) নং এবং (ii) নং বর্গ করে যোগ করে পাই,

$4{\cos }^2\frac{\alpha -\beta }{2}({\sin }^2\frac{\alpha +\beta }{2}+{\cos }^2\frac{\alpha +\beta }{2})=s^2+t^2$

বা, $4{\cos }^2\frac{\alpha -\beta }{2}=s^2+t^2$

বা, ${\cos }^2\frac{\alpha -\beta }{2}=\frac{s^2+t^2}{4}$

বা, ${\sec }^2\frac{\alpha -\beta }{2}=\frac{4}{s^2+t^2}$

বা, $1+{\tan }^2\frac{\alpha -\beta }{2}=\frac{4}{s^2+t^2}$

বা, ${\tan }^2\frac{\alpha -\beta }{2}=\frac{4}{s^2+t^2}-1=\frac{4-s^2-t^2}{s^2+t^2}$

$\therefore \tan \frac{\alpha -\beta }{2}=\pm \sqrt{\frac{4-s^2-t^2}{s^2+t^2}}$, (প্রমাণিত)