Alternatively, You can use Eulers relation which is often handy when considering products of trig functions. Eulers formula states that the complex exponential e^ix = cox(x) + i*sin(x). This formula can be solved for sin and cos to get the following identities:

cos(x) = ( e^ix + e^-ix)/2

sin(x) = (e^ix - e^-ix)/2i

So, the solution becomes:

6*cosx*sinx = 6*( e^ix + e^-ix)*(e^ix - e^-ix)/4i

= 3(e^ix * e^ix + e^ix * e^-ix - e^ix * e^-ix - e^-ix * e^-ix) / 2i

= 3(e^i2x - e^-i2x)/2i

= 3 sin(2x)

We have to express 6*(sin x)*(cos x) as a single sine or cosine function.

Now we know that sin 2x = 2*(sin x)*(cos x).

We use this relation here.

6 sin x cos x

=> 3*2*(sin x)*(cos x)

=> 3*sin 2x

**Therefore 6*(sin x)*(cos x) = 3*sin 2x.**

The question put is to express the 6sinxcosx as the function of single sine of cosine function.

As it is, the the given finction a single function with both sinx and cosx. f(x) = 6sinxcosx.

We can rewrite the function as a function of sinx as below:

f(sinx) = 6 sinx(1-sin^2x)^(1/2), as cosx = (1-sin^2x).

We can write the function purely as a function of K*sinX as below:

6sinxcosx = 3 (2sinxcosx) = 3sin(2x)

6 sinxcosx = 3sin(2x) , as sin2x= sin(x+x) = sinxcosx+cosxsinx = 2sinxcosx.

We can write 6sinxcosx = sin(2x) = cos(90-2x)

We cane write the 6sinxcosx = 6cosx*(1-cos2^)^(1/2).