How do you use the product to sum formulas to write \[6\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right)\] as a sum or difference?

Hint: In order to find the solution of the given question that is to use the product to sum formulas to write \[6\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right)\] as a sum or difference apply one of the identities of trigonometry that is named as product of sum formula represented as \[2\sin \left( x \right)\cos \left( y \right)=\sin \left( x+y \right)+\sin \left( x-y \right)\]. \[\sin \left( 2x \right)=2\sin \left( x \right)\cos \left( x \right)\] and solve the given expression further to get the simplified answer.

Complete step by step answer:
According to the question, given expression in the question is as follows:
\[6\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right)\]
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)\]
Applying one of the identities of trigonometry which is named as product of sum formula that is \[2\sin \left( x \right)\cos \left( y \right)=\sin \left( x+y \right)+\sin \left( x-y \right)\] in the above expression we get:
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)=3\left( \sin \left( \dfrac{\pi }{4}+\dfrac{\pi }{4} \right)+\sin \left( \dfrac{\pi }{4}-\dfrac{\pi }{4} \right) \right)\]
Now simplify the above expression, by solving the bracket outside the above expression, we get:
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)=3\sin \left( \dfrac{\pi }{4}+\dfrac{\pi }{4} \right)+3\sin \left( \dfrac{\pi }{4}-\dfrac{\pi }{4} \right)\]
After this simplify the above expression, by solving the bracket of the angle of sine we will have:
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)=3\sin \left( \dfrac{\pi }{2} \right)+3\sin 0\]
We know that \[\sin \left( 0 \right)=0\], so applying this result in the above formula we will have:
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)=3\sin \left( \dfrac{\pi }{2} \right)\]
Therefore, the given expression \[6\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right)\] is equal to \[3\sin \left( \dfrac{\pi }{2} \right)\].

Note:
There’s an alternative way to solve the above question, which is as follows:
Given expression in the question is as follows:
\[6\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right)\]
We can rewrite the above expression as follows:
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)\]
Applying one of the identities of trigonometry that is \[\sin \left( 2x \right)=2\sin \left( x \right)\cos \left( x \right)\] in the above expression we get:
\[\Rightarrow 3\left( 2\sin \left( 2\cdot \dfrac{\pi }{4} \right) \right)\]
Now simplify the above expression, by solving the bracket of the angle of sine and the bracket outside the above expression, we get:
\[\Rightarrow 3\sin \left( \dfrac{\pi }{2} \right)\]
Hence, we can write the final answer as:
\[\Rightarrow 3\left( 2\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right) \right)=3\sin \left( \dfrac{\pi }{2} \right)\]
Therefore, the given expression \[6\sin \left( \dfrac{\pi }{4} \right)\cos \left( \dfrac{\pi }{4} \right)\] is equal to \[3\sin \left( \dfrac{\pi }{2} \right)\].