Question

If \[A + B + C = 180^\circ \], \[\dfrac{{\sin 2A + \sin 2B + \sin 2C}}{{\sin A + \sin B + \sin C}} = k\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\], then the value of \[k\] is equal to

Answer 1

Hint: Here we are asked to find the value of an unknown variable \[k\] by using the given data. We will find the value of that unknown variable by simplifying the given equation using trigonometric identities. We can also find or simplify the expression on one side of the equation then we can substitute it in that.
Formula: Some formulae that we need to know:
\[\sin 2\theta = 2\sin \theta \cos \theta \]
\[\sin 2A + \sin 2B = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right)\]
\[\sin \left( {180^\circ - \theta } \right) = \sin \theta \]
\[\sin \left( {90^\circ - \theta } \right) = \cos \theta \]
\[\cos \left( {180^\circ - \theta } \right) = - \cos \theta \]
\[\cos \left( {A - B} \right) - \cos \left( {A + B} \right) = 2\sin A\sin B\]
\[\cos \left( {A - B} \right) + \cos \left( {A + B} \right) = 2\cos A\cos B\]
\[\sin \theta = 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)\]

Complete step by step answer:
It is given that \[A + B + C = 180^\circ \] and \[\dfrac{{\sin 2A + \sin 2B + \sin 2C}}{{\sin A + \sin B + \sin C}}\] we aim to find the value of the unknown variable \[k\].
We will first consider the left-hand side of the given equation and simplify it.
Using the formula \[\sin 2A + \sin 2B = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right)\] with the angles as \[A,B\] in the numerator of the left-hand side expression we get
\[\dfrac{{\sin 2A + \sin 2B + \sin 2C}}{{\sin A + \sin B + \sin C}} = \dfrac{{2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + \sin 2C}}{{\sin A + \sin B + \sin C}}\]
Now using the formula \[\sin 2\theta = 2\sin \theta \cos \theta \] on the third term of the numerator we get
\[ = \dfrac{{2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + 2\sin C\cos C}}{{\sin A + \sin B + \sin C}}\]
Using the same formulae, that is \[\sin 2A + \sin 2B = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right)\] and \[\sin 2\theta = 2\sin \theta \cos \theta \] in the denominator we get
\[ = \dfrac{{2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + 2\sin C\cos C}}{{2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}}}\]
We are given that \[A + B + C = 180^\circ \] from this we get \[A + B = 180^\circ - C\] substituting this to the sine function in the first term of the numerator and denominator we get
\[ = \dfrac{{2\sin \left( {180^\circ - C} \right)\cos \left( {A - B} \right) + 2\sin C\cos C}}{{2\sin \left( {90^\circ - \dfrac{C}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}}}\]
We know that \[\sin \left( {180^\circ - \theta } \right) = \sin \theta \] by using this formula with the angle \[C\] and using another formula \[\sin \left( {90^\circ - \theta } \right) = \cos \theta \] with the angle \[\dfrac{C}{2}\] we get
\[ = \dfrac{{2\sin C.\cos \left( {A - B} \right) + 2\sin C\cos C}}{{2\cos \dfrac{C}{2}.\cos \left( {\dfrac{{A - B}}{2}} \right) + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}}}\]
On simplifying the above, we get
\[ = \dfrac{{2\sin C\left[ {\cos \left( {A - B} \right) + \cos C} \right]}}{{2\cos \dfrac{C}{2}\left[ {\cos \left( {\dfrac{{A - B}}{2}} \right) + \sin \dfrac{C}{2}} \right]}}\]
Again, using that \[A + B = 180^\circ - C\] and \[\sin \left( {90^\circ - \theta } \right) = \cos \theta \] with the angle \[\dfrac{C}{2}\] we get
\[ = \dfrac{{2\sin C\left[ {\cos \left( {A - B} \right) + \cos \left( {180^\circ - \left( {A + B} \right)} \right)} \right]}}{{2\cos \dfrac{C}{2}\left[ {\cos \left( {\dfrac{{A - B}}{2}} \right) + \sin \left( {90^\circ - \left( {\dfrac{{A + B}}{2}} \right)} \right)} \right]}}\]
On simplifying this we get
\[ = \dfrac{{2\sin C\left[ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right]}}{{2\cos \dfrac{C}{2}\left[ {\cos \left( {\dfrac{{A - B}}{2}} \right) + \cos \left( {\dfrac{{A + B}}{2}} \right)} \right]}}\]
Using the formula \[\cos \left( {A - B} \right) - \cos \left( {A + B} \right) = 2\sin A\sin B\] and \[\cos \left( {A - B} \right) + \cos \left( {A + B} \right) = 2\cos A\cos B\] with the angles \[\dfrac{A}{2},\dfrac{B}{2},\dfrac{C}{2}\] we get
\[ = \dfrac{{2\sin C.2\sin A\sin B}}{{2\cos \dfrac{C}{2}.2\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
On further simplification we get
\[ = \dfrac{{\sin C\sin A\sin B}}{{\cos \dfrac{C}{2}\cos \dfrac{A}{2}\cos \dfrac{B}{2}}}\]
Now using the formula \[\sin \theta = 2\sin \left( {\dfrac{\theta }{2}} \right)\cos \left( {\dfrac{\theta }{2}} \right)\] with the angles as \[A,B,C\] we get
\[ = \dfrac{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}.2\sin \dfrac{B}{2}\cos \dfrac{B}{2}.2\sin \dfrac{C}{2}\cos \dfrac{C}{2}}}{{\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}}}\]
On simplifying this we get
\[ = 8\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\]
Therefore, we have simplified the left-hand side of the given equation. Let us substitute it in that equation.
\[8\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2} = k\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}\]
On simplifying this we get
\[k = 8\]
Thus, we got the value of the unknown variable.

Note:
 It is given that \[A + B + C = 180^\circ \] which implies \[\dfrac{{A + B + C}}{2} = \dfrac{{180^\circ }}{2}\], simplifying this for the value of \[\dfrac{C}{2}\] we get \[\dfrac{C}{2} = 90^\circ - \dfrac{{A + B}}{2}\]also simplifying it for the value of \[\dfrac{{A + B}}{2}\] we get \[\dfrac{{A + B}}{2} = 90^\circ - \dfrac{C}{2}\]. These derivatives are used in the above calculation.