Solveeit Logo
Login

Question

Question: \(\frac{\sin 2A + \sin 2B - \sin 2C}{\sin A + \sin B - \sin C}\) equal to...

sin⁡2A+sin⁡2B−sin⁡2Csin⁡A+sin⁡B−sin⁡C\frac{\sin 2A + \sin 2B - \sin 2C}{\sin A + \sin B - \sin C} equal to

A

cos⁡Acos⁡Bsin⁡Csin⁡A2sin⁡B2cos⁡C2\frac{\cos A\cos B\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}}

B

sin⁡Asin⁡Bcos⁡Ccos⁡A2cos⁡B2sin⁡C2\frac{\sin A\sin B\cos C}{\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}}

C

−cos⁡Acos⁡Bsin⁡Csin⁡A2sin⁡B2cos⁡C2- \frac{\cos A\cos B\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}}

D

−sin⁡Asin⁡Bcos⁡Ccos⁡A2cos⁡B2sin⁡C2- \frac{\sin A\sin B\cos C}{\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}}

Answer

cos⁡Acos⁡Bsin⁡Csin⁡A2sin⁡B2cos⁡C2\frac{\cos A\cos B\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}}

Explanation

Solution

(sin⁡2A+sin⁡2B)−sin⁡2C(sin⁡A+sin⁡B)−sin⁡C\frac{(\sin 2A + \sin 2B) - \sin 2C}{(\sin A + \sin B) - \sin C}=2sin⁡(A+B)cos⁡(A−B)−sin⁡2C2sin⁡(A+B2)cos⁡(A−B2)−sin⁡C\frac{2\sin(A + B)\cos(A - B) - \sin 2C}{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) - \sin C}

=2sin⁡Ccos⁡(A−B)−2sin⁡Ccos⁡C2sin⁡(π−C2)cos⁡(A−B2)−2sin⁡C2cos⁡C2\frac{2\sin C\cos(A - B) - 2\sin C\cos C}{2\sin\left( \frac{\pi - C}{2} \right)\cos\left( \frac{A - B}{2} \right) - 2\sin\frac{C}{2}\cos\frac{C}{2}}

=2sin⁡C[cos⁡(A−B)−cos⁡C]2cos⁡C2[cos⁡(A2−B2)−cos⁡(A2+B2)]\frac { 2 \sin C [ \cos ( A - B ) - \cos C ] } { 2 \cos \frac { C } { 2 } \left[ \cos \left( \frac { A } { 2 } - \frac { B } { 2 } \right) - \cos \left( \frac { A } { 2 } + \frac { B } { 2 } \right) \right] } [∵sin⁡C=2sin⁡C2cos⁡C2sin⁡C/2=sin⁡(π2−A+B2)=cos⁡(A+B)2]\begin{bmatrix} \because\sin C = 2\sin\frac{C}{2}\cos\frac{C}{2} \\ \sin C/2 = \sin\left( \frac{\pi}{2} - \frac{A + B}{2} \right) = \cos\frac{(A + B)}{2} \end{bmatrix}

=2sin⁡C[cos⁡(A−B)+cos⁡(A+B)]2cos⁡C2[cos⁡(A2−B2)−cos⁡(A2+B2)]\frac{2\sin C\lbrack\cos(A - B) + \cos(A + B)\rbrack}{2\cos\frac{C}{2}\left\lbrack \cos\left( \frac{A}{2} - \frac{B}{2} \right) - \cos\left( \frac{A}{2} + \frac{B}{2} \right) \right\rbrack}

=2sin⁡C[2cos⁡Acos⁡B]2cos⁡C2[2sin⁡A2sin⁡B2]\frac{2\sin C\lbrack 2\cos A\cos B\rbrack}{2\cos\frac{C}{2}\left\lbrack 2\sin\frac{A}{2}\sin\frac{B}{2} \right\rbrack} = cos⁡Acos⁡Bsin⁡Csin⁡A2sin⁡B2cos⁡C2\frac{\cos A\cos B\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}}.

Trick : ∵\mathbf{\because} sin⁡2A+sin⁡2B−sin⁡2C=4cos⁡Acos⁡Bsin⁡C\sin 2A + \sin 2B - \sin 2C = 4\cos A\cos B\sin Cand

sin⁡A+sin⁡B−sin⁡C=4sin⁡A2sin⁡B2cos⁡C2\sin A + \sin B - \sin C = 4\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}

⇒ sin⁡2A+sin⁡2B−sin⁡2Csin⁡A+sin⁡B−sin⁡C=cos⁡Acos⁡Bsin⁡Csin⁡A2sin⁡B2cos⁡C2\frac{\sin 2A + \sin 2B - \sin 2C}{\sin A + \sin B - \sin C} = \frac{\cos A\cos B\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2}}.