4sinacosacos2a=sin4a.

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To solve this equation, we can use the double angle identity for sine:

sin(2x) = 2sin(x)cos(x)

Substitute 2a in place of x:

sin(4a) = 2sin(2a)cos(2a)

Now we need to find sin(2a) and cos(2a):

sin(2a) = 2sin(a)cos(a)

cos(2a) = 2cos^2(a) - 1

Now substitute sin(2a) and cos(2a) back into the equation:

sin(4a) = 2(2sin(a)cos(a))(2cos^2(a) - 1)
sin(4a) = 4sin(a)cos(a)(2cos^2(a) - 1)
sin(4a) = 8cos(a)sin(a)cos^2(a) - 4sin(a)cos(a)
sin(4a) = 4cos(a)sin(a)
sin(4a) = sin(2a)

Therefore, 4sin(a)cos(a)cos(2a) = sin(4a) as required.