Question

Choose from the given alternative: In any triangle ABC which of the following is equivalent to$\sin 2A+\sin 2B+\sin 2C$ ?
(a) $4\sin A\sin B\sin C$
(b) $4\cos A\cos B\cos C$
(c) $\dfrac{3}{2}\sin A\cos B$
(d) $4\cos A\sin B\cos C$

Answer 1

Hint:
Use the formula of summation of two sine terms.
$\begin{aligned} & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\\ \end{aligned}$
Use angle sum property of triangle:
In any triangle $ABC$, $\angle A+\angle B+\angle C={{180}^{{}^\circ }}$. It is simply written as
A + B + C = 180°
Use basic formula of trigonometry like
$\begin{aligned} & \sin 2\theta =2\sin \theta \cos \theta \\\ & \sin ({{90}^{\circ }}-A)=\cos A \\\ & \sin ({{180}^{{}^\circ }}-A)=\sin A \\\ & \cos ({{90}^{\circ }}-A)=\sin A \\\ \end{aligned}$

Complete step-by-step answer:
We know that
$\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Using the formula mentioned above with C = 2A and D = 2B we get our expression converted as:
$\begin{aligned} & \sin 2A+\sin 2B+\sin 2C=\left( \sin 2A+\sin 2B \right)+\sin 2C \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)+\sin 2C \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin \left( A+B \right)\cos \left( A-B \right)+\sin 2C\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot (\text{ii)} \\\ \end{aligned}$

We know that in triangle ABC,
A + B + C = 180°
⇒ A + B = 180° - C
$\Rightarrow \sin \left( A+B \right)=\sin \left( {{180}^{{}^\circ }}-C \right)\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(iii)}$
Also, for any angle $\theta$ we have,
$\sin \theta =\sin \left( {{180}^{{}^\circ }}-\theta \right)$

Using the above formula in equation (iii) we get
$\sin \left( A+B \right)=\sin \left( {{180}^{{}^\circ }}-C \right)=\sin C\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(iv)}$
For any angle $\theta$ we have
$\sin 2\theta =2\sin \theta \cos \theta$
Using this with $\theta =C$ we get,
$\sin 2C=2\sin C\cos C\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(v)}$
Using equation (iv) and equation (v) in equation (ii) we get,
$\begin{aligned} & \sin 2A+\sin 2B+\sin 2C=2\sin \left( A+B \right)\cos \left( A-B \right)+\sin 2C \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin C\cos \left( A-B \right)+2\sin C\cos C \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin C\left( \cos \left( A-B \right)+\cos C \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(vi)} \\\ \end{aligned}$

We know that for any angle C and D,
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Using the above formula in equation (vi) we get
$\begin{aligned} & \sin 2A+\sin 2B+\sin 2C\,=2\sin C\left( \cos \left( A-B \right)+\cos C \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sin C\left( 2\cos \left( \dfrac{A-B+C}{2} \right)\cos \left( \dfrac{A-B-C}{2} \right) \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=4\sin C\cos \left( \dfrac{A-B+C}{2} \right)\cos \left( \dfrac{A-B-C}{2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(vii)} \\\ \end{aligned}$

We have
A + B + C = 180°
⇒A – B + C = 180°-2B
$\begin{aligned} & \Rightarrow \dfrac{A-B+C}{2}=\dfrac{{{180}^{{}^\circ }}-2B}{2}={{90}^{\circ }}-B \\\ & \Rightarrow \cos \left( \dfrac{A-B+C}{2} \right)=\cos \left( {{90}^{\circ }}-B \right)\,\,\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(viii)} \\\ \end{aligned}$

We know for any angle $\theta$,
$\cos \left( {{90}^{{}^\circ }}-\theta \right)=\sin \theta$
Applying this formula in the equation (viii) we get
$\cos \left( \dfrac{A-B+C}{2} \right)=\cos \left( {{90}^{\circ }}-B \right)=\sin B\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(ix)}$
Again,
A – B – C = 90° - 2B – 2C

& \Rightarrow \dfrac{A-B-C}{2}=\dfrac{{{180}^{{}^\circ }}-2B-2C}{2}={{90}^{\circ }}-B-C \\\ & \Rightarrow \cos \left( \dfrac{A-B+C}{2} \right)=\cos \left( {{90}^{\circ }}-B-C \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sin \left( B+C \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sin \left( {{180}^{{}^\circ }}-A \right)\,\, \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sin \left( A \right)\,\,\,\,\,\cdot \cdot \cdot \text{(x)} \\\ \end{aligned}$$ Using equation (ix) and equation (x) in equation (vii) we get $\begin{aligned} & \sin 2A+\sin 2B+\sin 2C\,=4\sin C\cos \left( \dfrac{A-B+C}{2} \right)\cos \left( \dfrac{A-B-C}{2} \right) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=4\sin C\sin B\sin A \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=4\sin A\sin B\sin C \\\ \end{aligned}$ Hence option (a) is correct. Note: There are lots of trigonometric formulas used. There is a chance of making mistakes in applying those formulas. You can also start taking another two sine terms in a group and then converting them into products of sine and cosine. That will also yield the same result.