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Show that 2sin(2x)-3cos(2x)-3sin(x)+3=sin(x)(4cos(x)+6sin(x)-3)

When proving trigonometric identites, we must show that the left hand side of the equation = the right hand side. Here we will start with the left hand side (LHS) and show that it is equivalent to the right hand side (RHS).

LHS=2sin(2x)-3cos(2x)-3sin(x)+3

Using the double angle rules for sin(2x) and cos(2x);

LHS=2(2sin(x)cos(x))-3(cos2(x)-sin2(x))-3sin(x)+3

Notice that the RHS has sin(x) factorised out, meaning that every term in the LHS has a common factor of sin(x). Currently the LHS has a cos2x term, but we can change this to a sin2x term using the identity: cos2(x)=1-sin2(x) 

LHS=2(2cos(x)sin(x))-3(1-sin2(x)-sin2(x))-3sin(x)+3

=4cos(x)sin(x)-3(1-2sin2(x))-3sin(x)+3

=4cos(x)sin(x)-3+6sin2(x)-3sin(x)+3

=4cos(x)sin(x)+6sin2(x)-3sin(x)

=sin(x)(4cos(x)+6sin(x)-3)

=RHS

We have shown that LHS=RHS, therefore the proof is complete.

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