সৃজনশীলঃ ‘দৃশ্যকল্প-১: M= cos25α+cos25β-sin25g+2cos5α…’ বরিশাল বোর্ড ২০২২ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$\alpha +\beta +\gamma =\pi$

$\Rightarrow 5\alpha +5\beta +5\gamma =5\pi$

এখানে, ${\cos }^{2}5\alpha +{\cos }^{2}5\beta -{\sin }^{2}5\gamma$

$\Rightarrow \frac{1}{2}(2{\cos }^{2}A+2{\cos }^{2}B-2{\sin }^{2}C$

$=\frac{1}{2}[1+\cos 2A-(1-\cos 2C\left)\right]+{\cos }^{2}B$

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

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

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

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

$=\cos (A+C)\cos (A-C)+{\cos }^{2}B$

$=\cos (\pi -B)\cos (A-C)+{\cos }^{2}B$

$=-\cos B-\cos (A-C)+{\cos }^{2}B$

$=-\cos B(\cos (A-C)-\cos B)$

$=-\cos B[\cos (A-C)-\cos (\pi -(A+C)\left)\right]$

$=-\cos B[\cos (A-C)+\cos (A+C\left)\right]$

$=-\cos B\times 2\cos A\cos C$

$=-2\cos A\cos B\cos C$

$=-2\cos 5\alpha \cos 5\beta \cos 5\gamma$

$\therefore M={\cos }^{2}5\alpha +{\mathrm{ces}}^{2}5\beta -{\sin }^{2}5\gamma +2\cos 5\alpha \times \cos 5\beta \cos 5\gamma$

$=-2\cos 5\alpha \cos 5\beta \cos 5\gamma +2\cos 5\alpha \cos 5\beta \cos 5\gamma$

$=0$

$\therefore M=0$ (proved)

L.S $=\frac{b-c}{2a}$

$=\frac{2R\sin B-2R\sin C}{2\times 2R\sin A}$.

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

$=\frac{2RC(\sin B-\sin C)}{RR2\sin A}$

$=\frac{2\cos \frac{B+C}{2}\sin \frac{B-C}{2}}{2\sin {60}^{\circ }}$

$=\frac{2\cos (\frac{\pi }{2}-\frac{A}{2})\sin \frac{B-C}{2}}{2\times \frac{\sqrt{3}}{2}}$

$=\frac{2\sin \frac{A}{2}\sin \frac{B-C}{2}}{\sqrt{3}}$

$=\frac{2}{\sqrt{3}}\times \sin {30}^{\circ }\times \sin \frac{B-C}{2}$

$=\frac{2}{\sqrt{3}}\times \frac{1}{2}\times \sin \frac{B-C}{2}$

$=\frac{1}{\sqrt{3}}\sin \frac{B-C}{2}=R\cdot S$ (Proved)