Question

2sin4tita.cos2tita Express the following as a sum or difference of two angle

Answer 1

Answer:

Sin6x  + Sin2x

Sin(4x + 2x)   + Sin(4x-2x)

Step-by-step explanation:

2Sin(4x).Cos(2x)

Using Sin(A+b) = SinACosB + CosASinB

Sin(4x) = Sin(3x+x) = Sin3xCosx + Cos3xSinx

Cos(A-B) = CosACosB + sinASinB

Cos2x = Cos(3x-x) = Cos3xCosx + Sin3xSinx

2Sin(4x).Cos(2x)

= 2 (Sin3xCosx + Cos3xSinx)(Cos3xCosx + Sin3xSinx)

= 2Sin3xCos3xCos²x + 2Sin²3xCosxSinx + 2Cos²3xSinxCosx + 2Cos3xSin3xSin²x)

= 2Sin3xCos3x(Cos²x + Sin²x)  + 2CosxSinx(Sin²3x + Cos²3x)

Using Cos²x + Sin²x = 1 and Sin²3x + Cos²3x = 1

= 2Sin3xCos3x + 2CosxSinx

Using Sin2x = 2SinxCosx

= Sin6x  + Sin2x

= Sin(4x + 2x)   + Sin(4x-2x)

Answer 2

Answer:

$sin\:6\theta+sin\:2\theta$

Step-by-step explanation:

Formula used:

$2\:sinA\:cosA=sin(A+B)+sin(A-B)$

Now,

$2\:sin4\theta\:cos2\theta\\\\=sin(4\theta+2\theta)+sin(4\theta-2\theta)\\\\=sin(6\theta)+sin(2\theta)$