Question

How do I prove \[2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)=\sin C+\sin D\]?

Answer 1

Hint: To solve the given question, we should know some trigonometric expansion formula. The trigonometric properties we will be using are \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\], and the other one is \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\]. As this is a property, we can use it in both directions. That is we should also know that, we can say that\[\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)\], and \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\].

Complete step-by-step solution:
We are asked to prove the statement \[2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)=\sin C+\sin D\]. To prove a statement we need show that one side of the statement either left or right can be expressed as the other side. Let’s take the left-hand side of the given statement,
\[LHS=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]
This can also be written as,
\[\Rightarrow \sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)+\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)\]
Adding and subtracting \[\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)\] from the above expression we get
\[\begin{align}
  & \Rightarrow \sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)+\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
 & +\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right)-\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right) \\
\end{align}\]
Rearranging the terms in above expression, we get
\[\begin{align}
  & \Rightarrow \sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)-\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right) \\
 & +\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)+\sin \left( \dfrac{C-D}{2} \right)\cos \left( \dfrac{C+D}{2} \right) \\
\end{align}\]
We know the trigonometric property which states that, \[\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)\]one the third and fourth term, and \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\] on the first and second term of the of the expression. It can be written as,
\[\Rightarrow \sin \left( \dfrac{C+D}{2}-\dfrac{C-D}{2} \right)+\sin \left( \dfrac{C+D}{2}+\dfrac{C-D}{2} \right)\]
Simplifying the arguments of the above trigonometric functions we get,
\[\Rightarrow \sin D+\sin C\]
The right-hand side of the given question is \[\sin C+\sin D\]. We have shown that the left side of the expression can be expressed as the right side.
Hence, as \[LHS=RHS\]. The given statement has been proved.

Note: The statement given in the question should be remembered as it might be useful to solve different questions of trigonometry. It can also be used to prove other statements. We can also find a property for the trigonometric function cosine, using a similar method to prove.