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The expression on the left needs to be reduced to terms involving sin(x) or cos(x).

sin(2x)≡2sin(x)cos(x); cos(2x)≡1-2sin2(x) or 2cos2(x)-1.

5cos(x)-2sin(2x)≡5cos(x)-4sin(x)cos(x)≡cos(x)(5-4sin(x)).

2cos(2x)-3sin(x)+8≡2-4sin2(x)-3sin(x)+8=10-3sin(x)-4sin2(x)=(2+sin(x))(5-4sin(x)).

(The clue to the factorisation is the appearance of sin(x)+2, the same as 2+sin(x), in the denominator on the RHS.)

So the expression becomes:

(cos(x)(5-4sin(x)))/((2+sin(x))(5-4sin(x)))=cos(x)/(2+sin(x)) because the common factor 5-4sin(x) cancels out. QED

Note that 5-4sin(x) can never be zero because sin(x) cannot exceed 1. This implies that it is safe to cancel the factor because the identity is true for all x.

by Top Rated User (1.2m points)

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