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

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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).


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


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) 







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

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