Question

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?

Answer 1

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 $3\sin \left( \dfrac{\pi }{2} \right)$.