সৃজনশীলঃ ‘f(x)−cos⁡x,g(x)=sin⁡xএবংt(x)−tan⁡x…’ নটরডেম কলেজ ২০২৪ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া आছে, $f\left(x\right)=\cos x$

$f\left(x\right)=\cos A,f\left(B\right)=\cos B,f\left(C\right)=\cos C$

এখন, $\left\{f\right(A){\}}^2+\{f\left(B\right){\}}^2-\left\{f\right(C){\}}^2$

বা, ${\cos }^{2}A+{\cos }^{2}B-{\cos }^{2}C=2P$

বা,$2P=\frac{1}{2}\{2{\cos }^{2}A+2{\cos }^{2}B\}-{\cos }^{2}C$

বা, $2P=\frac{1}{2}\{1+\cos 2A+1+\cos 2B\}-{\cos }^{2}C$

বা, $2P=1+\frac{1}{2}(\cos 2A+\cos 2B)-{\cos }^{2}C$

বা, $2P=1+\frac{1}{2}\cdot 2\cos (A+B)\cos (A-B)-{\cos }^{2}C$

বা, $2P=1+\cos (\frac{\pi }{2}-C)\cos (A-B)-{\cos }^{2}C[\because A+B+C=\frac{\pi }{2}\}$

বা, $2P=1+\sin C\cos (A-B)-(1-{\sin }^{2}C)$

বা, $2P=1\sin C\{\cos (A-B)+\sin C)-1$

বা, $2P=\sin C\{\cos (A-B)+\sin [\frac{\pi }{2}-(A+B\left)\right]\}[\because A+B+C=\frac{\pi }{2}$

বা, $2P=\sin C\{\cos (A-B)+\cos (A+B\left)\right\}$

বা, $2P=2\cos A\cos B\sin C$

$\therefore P=\cos A\cos B\sin C$

নির্ণেয় মান : P $=\cos A\cos B\cos C$

দেওয়া আছে, $f\left(x\right)=\cos x$

বামপক্ষ

$=a^{3}f\{\frac{\pi }{2}-(B-C\left)\right\}+b^{3}f\{\frac{\pi }{2}-(C-A\left)\right\}+c^{3}f\{\frac{\pi }{2}-(A-B\left)\right\}$

$=a^3\cos \{\frac{\pi }{2}-(B-C\left)\right\}+b^3\cos \{\frac{\pi }{2}-(C-A\left)\right\}+c^3\cos \{\frac{\pi }{2}-(A-B\left)\right\}$

$=a^3\sin (B-C)+b^3\sin (C-A)+c^3\sin (A-B)$

$=a^2\cdot 2R\sin A\cdot \sin (B-C)+b^2\cdot 2R\sin B\cdot \sin (C-A)+c^2\cdot 2R\sin C\cdot \sin (A-B)$

$=2a^{2}R\sin \{\pi -(B+C\left)\right\}\sin (B-C)+2b^{2}R\sin \{\pi -(C+A\left)\right\}\sin (C-A)+2c^{2}R\sin \{\pi -(A+B\left)\right\}\sin (A-B)$

ง$[\because △\mathrm{ABC}งA+B+C=\pi ]$

$=2a^{2}R\sin (B+C)\sin (B-C)+2b^{2}R\sin (C+A)\sin (C-A)+2c^{2}R\sin (A+B)\sin (A-B)$

$=2a^{2}R({\sin }^{2}B-{\sin }^{2}C)+2b^{2}R({\sin }^{2}C-{\sin }^{2}A)+2c^2({\sin }^{2}A-{\sin }^{2}B)$

$=2a^{2}R(\frac{b^2}{4R^2}-\frac{c^2}{4R^2})+2b^{2}R(\frac{c^2}{4R^2}-\frac{a^2}{4R^2})+2c^{2}R(\frac{a^2}{4R^2}-\frac{b^2}{4R^2})$

$[\because △ABC-a\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R]$

$=\frac{a^2{b}^2}{2R}-\frac{a^2{c}^2}{2R}+\frac{b^2{c}^2}{2R}-\frac{a^2{b}^2}{2R}+\frac{a^2{c}^2}{2R}-\frac{\sin b^2{c}^2}{2R}$

 = 0 = ডানপক্ষ

$\therefore a^{3}f\{\frac{\pi }{2}-(B-C\left)\right\}+b^{3}f\{\frac{\pi }{2}-(C-A\left)\right\}+c^{3}f(\frac{\pi }{2}-(A-B\left)\right\}=0$ (প্রমাণিত)