Solve the trigonometric expression \[\sin A+\sin 3A+\sin 5A+\sin 7A=\]

Hint: In this type of question we have to use the concept of trigonometry. Here, we will use the formula \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] and apply it to \[\left( \sin A+\operatorname{Sin}3A \right)\] and \[\left( \sin 5A+\sin 7A \right)\] separately. Also we have to use \[\cos \left( -\theta \right)=\cos \theta \]. Then we can simplify and obtain the result.

Complete step-by-step solution:
Now, we have to solve \[\sin A+\sin 3A+\sin 5A+\sin 7A\]
For this let us consider,
\[\Rightarrow \left( \sin A+\operatorname{Sin}3A \right)+\left( \sin 5A+\sin 7A \right)\]
As we know that, \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] we can simplify both the brackets
\[\Rightarrow \left[ 2\sin \left( \dfrac{A+3A}{2} \right)\cos \left( \dfrac{A-3A}{2} \right) \right]+\left[ 2\sin \left( \dfrac{5A+7A}{2} \right)\cos \left( \dfrac{5A-7A}{2} \right) \right]\]
By simplifying it further we can write,
\[\Rightarrow \left[ 2\sin \left( 2A \right)\cos \left( -A \right) \right]+\left[ 2\sin \left( 6A \right)\cos \left( -A \right) \right]\]
Now, as we know that, \[\cos \left( -\theta \right)=\cos \theta \] we get, \[\cos \left( -A \right)=\cos A\] and hence the above expression becomes,
\[\begin{align}
  & \Rightarrow 2\sin 2A\cos A+2\sin 6A\cos A \\
 & \Rightarrow 2\cos A\left( \sin 2A+\sin 6A \right) \\
\end{align}\]
Now by using the same formula that is \[\sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] we can write,
\[\begin{align}
  & \Rightarrow 2\cos A\left[ 2\sin \left( \dfrac{2A+6A}{2} \right)\cos \left( \dfrac{2A-6A}{2} \right) \right] \\
 & \Rightarrow 2\cos A\left[ 2\sin 4A\cos \left( -2A \right) \right] \\
\end{align}\]
Again by using \[\cos \left( -\theta \right)=\cos \theta \] we can write, \[\cos \left( -2A \right)=\cos 2A\] and hence we get,
\[\begin{align}
  & \Rightarrow 2\cos A\left[ 2\sin 4A\cos 2A \right] \\
 & \Rightarrow 4\cos A\cos 2A\sin 4A \\
\end{align}\]
Hence, the solution of the given equation is \[4\cos A\cos 2A\sin 4A\].
Thus, \[\sin A+\sin 3A+\sin 5A+\sin 7A=4\cos A\cos 2A\sin 4A\]

Note: In this type of question students have to remember the trigonometric properties. As in this question four terms of sine are present all with positive sign so to simplify the given expression one may use any of the combination of terms. That is, instead of a combination of first, second and third fourth terms one may use first third and second fourth terms together for simplification. Also instead of left to right one may choose terms from right to left for simplification i.e. one of the students may use the combination as \[\left( \sin 7A+\sin 5A \right)+\left( \sin 3A+\sin A \right)\]. In such a combination students always get a positive angle for cosine also and hence at that time no need to use the formula \[\cos \left( -\theta \right)=\cos \theta \]. Students have to take care during the simplification of terms.