Question

Express $\sin 3A+\sin 3B+\sin 3C$ as the product of three trigonometric ratios where $A, B, C$ are the angles of a triangle. If the given expression is zero, then at least one angle of the triangle is ${{60}^{\circ }}$.

Answer 1

Hint: For answering this question we will use the value of sum of angles in a triangle $A+B+C=\pi $ and simplify the expression we have $\sin 3A+\sin 3B+\sin 3C$ using different equations like $\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$ and$\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B$ after various steps the expression will be in the form of product of trigonometric ratios. For this we will find the solutions.

Complete step-by-step solution:
Here from the question, we know that $A, B, C$ are the angles of a triangle so we can say that $A+B+C=\pi $ because the sum of angles in a triangle is $\pi $.
So we can say that $A+B=\pi -C$ from $A+B+C=\pi $ .
By simplifying the expression we have $\sin 3A+\sin 3B+\sin 3C$ by using $\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$ we will have $2\sin \dfrac{3A+3B}{2}\cos \dfrac{3A-3B}{2}+\sin 3C$.
By expanding $\sin 3C$ using $\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$ we will have $2\sin \dfrac{3A+3B}{2}\cos \dfrac{3A-3B}{2}+2\sin \dfrac{3C}{2}\cos \dfrac{3C}{2}$ .
By substituting $A+B=\pi -C$ we will have $2\sin \dfrac{3\pi -3C}{2}\cos \dfrac{3A-3B}{2}+2\sin \dfrac{3C}{2}\cos \dfrac{3C}{2}$ .
As we know that $\sin \left( \dfrac{3\pi }{2}-x \right)=-\cos x$ we can simply write it as $-2\cos \dfrac{3C}{2}\cos \dfrac{3A-3B}{2}+2\sin \dfrac{3C}{2}\cos \dfrac{3C}{2}$ .
By taking $2\cos \dfrac{3C}{2}$ as common we will have $-2\cos \dfrac{3C}{2}\left( \cos \dfrac{3A-3B}{2}-\sin \dfrac{3C}{2} \right)$ .
By substituting $C=\pi -\left( A+B \right)$ we will have $-2\cos \dfrac{3C}{2}\left( \cos \dfrac{3A-3B}{2}-\sin \dfrac{3\left( \pi -A-B \right)}{2} \right)$ .
As we know that $\sin \left( \dfrac{3\pi }{2}-x \right)=-\cos x$ we can simply write it as $-2\cos \dfrac{3C}{2}\left( \cos \dfrac{3A-3B}{2}+\cos \dfrac{3\left( A+B \right)}{2} \right)$
As we know that $\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B$ we can simply write it as $-2\cos \dfrac{3C}{2}\left( 2\cos \dfrac{3A}{2}\cos \dfrac{3B}{2} \right)$ .
By simplifying it we will have $-4\cos \dfrac{3C}{2}\left( \cos \dfrac{3A}{2}\cos \dfrac{3B}{2} \right)$ .
Hence $\sin 3A+\sin 3B+\sin 3C=-4\cos \dfrac{3A}{2}\cos \dfrac{3B}{2}\cos \dfrac{3C}{2}$ .
Hence we can conclude that the expression $\sin 3A+\sin 3B+\sin 3C$ can be as the product of three trigonometric ratios $-4\cos \dfrac{3A}{2}\cos \dfrac{3B}{2}\cos \dfrac{3C}{2}$ where $A,B,C$ are the angles of a triangle.
If the given expression is zero, then we need to find the solution.
The expression we have is $\sin 3A+\sin 3B+\sin 3C=-4\cos \dfrac{3A}{2}\cos \dfrac{3B}{2}\cos \dfrac{3C}{2}$ .
When it is zero we will have $-4\cos \dfrac{3A}{2}\cos \dfrac{3B}{2}\cos \dfrac{3C}{2}=0$ . This implies that at least anyone’s trigonometric ratio should be zero. For $A$ this can be mathematically given as $\cos \dfrac{3A}{2}=0$.
As we know that solutions for $\cos \theta =0$ is ${{90}^{\circ }}$ . We can say that $\dfrac{3A}{2}={{90}^{\circ }}$ .
By simplifying this we will have $A={{60}^{\circ }}$.
Hence we can conclude that if the given expression is zero, then at least one angle of the triangle is ${{60}^{\circ }}$.

Note: While using the different equations we should be careful and sure with them. These formulae $\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B$ and $\cos \left( A+B\right)-\cos \left( A-B \right)=-2\sin A\sin B$ are obtained from $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ . For example if we made a mistake and had used $\cos \left( A+B \right)+\cos \left( A-B \right)=-2\sin A\sin B$ instead of $\cos \left( A+B \right)+\cos \left( A-B \right)=2\cos A\cos B$. We will have a conclusion as $\sin 3A+\sin 3B+\sin 3C=4\sin \dfrac{3A}{2}\sin \dfrac{3B}{2}\cos \dfrac{3C}{2}$. We will end up having a complete mess and unable to solve the question. We have many trigonometric formulae that we can use in sums like these they are $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$, $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ and we can say that $\sin \left( A+B \right)+\sin \left( A-B \right)=2\sin A\cos B$ and $\sin \left( A+B \right)-\sin \left( A-B \right)=2\cos A\sin B$ .