Answer:
[tex]\cos(2x)\sin(5x)=\dfrac{\sin (7x)+\sin (3x)}{2}[/tex]
Step-by-step explanation:
Given: [tex]\cos(2x)\sin(5x)[/tex]
Formula:
[tex]\sin A+\sin B=2\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})[/tex]
[tex]\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})=\dfrac{\sin A+\sin B}{2}[/tex]
Compare the given expression with formula
[tex]\cos(2x)\sin(5x)=\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2})=\dfrac{\sin A+\sin B}{2}[/tex]
Therefore,
[tex]\dfrac{A+B}{2}=5x\Rightarrow A+B=10x[/tex]
[tex]\dfrac{A-B}{2}=2x\Rightarrow A-B=4x[/tex]
Using two system of equation of A and B to solve for A and B
Add both equation to eliminate B
[tex]2A=14x[/tex]
[tex]A=7x[/tex]
Substitute A into A+B=10x
[tex]7x+B=10x[/tex]
[tex]B=3x[/tex]
Substitute A and B into formula
[tex]\cos(2x)\sin(5x)=\sin(\dfrac{7x+3x}{2})\cos(\dfrac{7x-3x}{2})=\dfrac{\sin (7x)+\sin (3x)}{2}[/tex]
Hence, Product as sum form [tex]\cos(2x)\sin(5x)=\dfrac{\sin (7x)+\sin (3x)}{2}[/tex]
