Question

A+B+C=π হলে প্রমান কর যে,sin(B+C-A)+sin(C+A-B)+(A+B-C)=4sinA.sinB.sinC
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Answer 1

Solution :

$\bf{sin(B + C - A) + sin(C + A - B) + sin (A + B - C) = 4sinAsinBsinC}$

From the above equation , we get :

• LHS = $\bf{sin(B + C - A) + sin(C + A - B) + sin (A + B - C)}$

• RHS = $\bf{4sinAsinBsinC}$

Given that ,

$\bf{A + B + C = \pi}$

From the above equation, we get :

$:\implies \bf{B + C = \pi - A}$⠀⠀⠀⠀⠀⠀⠀⠀⠀Eq.(i)

$:\implies \bf{A + C = \pi - B}$⠀⠀⠀⠀⠀⠀⠀⠀⠀Eq.(ii)

$:\implies \bf{A + B = \pi - C}$⠀⠀⠀⠀⠀⠀⠀⠀⠀Eq.(iii)

Now substituting them in the LHS of the equation , we get :

$:\implies \bf{sin(B + C - A) + sin(C + A - B) + sin (A + B - C)}$

$:\implies \bf{sin(\pi - A - A) + sin(\pi - B - B) + sin(\pi - C - C)}$

$:\implies \bf{sin(\pi - 2A) + sin(\pi - 2B) + sin(\pi - 2C)}$

$\boxed{\begin{minipage}{8 cm} \textsf{We know that :} \\ \\ :\implies \bf{sin(\pi - \theta) = sin\theta} \\ \\ \textbf{Now by using it in the equation , we get :} \end{minipage}}$

$:\implies \bf{sin2A + sin2B + sin2C}$

$\boxed{\begin{minipage}{8 cm} \textsf{We know that :} \\ \\ \bullet\quad :\implies \bf{sin2x + sin2y = 2sin(x + y)cos(x - y)} \\ \\ \bullet\quad :\implies \bf{sin2x = 2sinxcosx} \\ \\ \textbf{Now by using it in the equation , we get :} \end{minipage}}$

$:\implies \bf{2sin(A + B)cos(A - B) + sin2C}$

$:\implies \bf{2sin(A + B)cos(A - B) + 2sinCcosC}$

By taking 2sinC common in the equation , we get :

$:\implies \bf{2sinC[2cos(A - B) + 2cosC)]}$

$\boxed{\begin{minipage}{8 cm} \textsf{We know that :} \\ \\ \bullet\quad :\implies \bf{cosx + cosy = 2cos\bigg(\dfrac{x + y}{2}\bigg)cos\bigg(\dfrac{x - y}{2}\bigg)} \\ \\ \textbf{Now by using it in the equation , we get :} \end{minipage}}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{A - B + C}{2}\bigg)cos\bigg(\dfrac{A - B - C}{2}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - B - B}{2}\bigg)cos\bigg(\dfrac{A- (B + C)}{2}\bigg)\bigg]} \\ \\ \bf{[\because \bullet\:A + C = \pi - B \\ \\ \bullet\:B + C = \pi - A)]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - 2B}{2}\bigg)cos\bigg(\dfrac{A- (\pi - A)}{2}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - 2B}{2}\bigg)cos\bigg(\dfrac{A- \pi + A}{2}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - 2B}{2}\bigg)cos\bigg(\dfrac{- \pi + 2A}{2}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - 2B}{2}\bigg)cos\bigg(\dfrac{- (\pi - 2A)}{2}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - 2B}{2}\bigg)cos\bigg\{-\bigg(\dfrac{\pi - 2A}{2}\bigg)\bigg\}\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi - 2B}{2}\bigg)cos\bigg(\dfrac{\pi - 2A}{2}\bigg)\bigg]} \quad \bf{\because cos(-x) = cosx}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi}{2} - \dfrac{2B}{2}\bigg)cos\bigg(\dfrac{\pi}{2} - \dfrac{2A}{2}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{\pi}{2} - \dfrac{\not{2}B}{\not{2}}\bigg)cos\bigg(\dfrac{\pi}{2} - \dfrac{\not{2}A}{\not{2}}\bigg)\bigg]}$

$:\implies \bf{2sinC\bigg[2cos\bigg(\dfrac{180^{\circ}}{2} - B\bigg)cos\bigg(\dfrac{180^{\circ}}{2} - A\bigg)\bigg]}\quad \bf{\because \pi = 180^{\circ}}$

$:\implies \bf{4sinCcos(90^{\circ} - Bcos)(90^{\circ} - A)}$

$:\implies \bf{4sinCsinBsinA}$

$\boxed{\therefore \bf{4sinAsinBsinC}}$

Hence the LHS of the Equation is $\bf{4sinCsinBsinA}$

Now by putting the LHS and RHS together , we get :

$:\implies \bf{4sinAsinBsinC = 4sinAsinBsinC}$

$\boxed{\therefore \bf{sin(B + C - A) + sin(C + A - B) + sin (A + B - C) = 4sinAsinBsinC}}$

Hence LHS = RHS.

Proved !