Question

2 sin 4x cos 2x sum or difference of two trigonometric function​

Answer 1

Answer:

$\displaystyle{2\sin4x \ \cos2x}\\\\\displaystyle{\implies\sin6x\right)+\sin2x}$

Step-by-step explanation:

$\displaystyle{2\sin4x \ \cos2x}\\\\\\\displaystyle{Using \ formula}\\\\\\\displaystyle{2\sin A\cos B=\sin(A+B)+\sin(A-B)}\\\\\\\displaystyle{2\sin4x \ \cos2x}\\\\\\\displaystyle{\implies\sin\left(4x+2x\right)+\sin\left(4x-2x\right)}\\\\\\\displaystyle{\implies\sin\left(6x\right)+\sin\left(2x\right)}$

$\displaystyle{\implies\sin6x\right)+\sin2x}$

Thus we get answer .

Answer 2

Answer:

Question :

2 sin 4x cos 2x sum of difference of two trigonometry function .

To find :

The sum or difference of two trigonometry function .

Formula :

2 sin A and B = sin (A + B) + sin (A - B)

= 2 sin 4x cos 2x

= sin (4x + 2x) + sin(4x - 2x)

= sin (6x) + sin (2x )

= sin 6x + sin 2x .