How do you verify # (3sinx-2cos^2x)/(2+sinx) = 2sinx -1#?

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Answer 1 · 2015-11-05T13:07:51

Multiply both terms by the denominator:

#cancel((2+sin(x)))(3sin(x)-2cos^2(x))/cancel((2+sin(x))) = (2sin(x)-1)(2+sin(x))#

So, we have

#3sin(x)-2cos^2(x) = 4sin(x)+2sin^2(x)-2-sin(x)#

We have #3sin(x)# at the left and #4sin(x)-sin(x)# at the right, so they cancel, leaving

#-2cos^2(x)=2sin^2(x)-2#

Divide everything by #2#:

#-cos^2(x) = sin^2(x)-1#

Rearrange and obtain the fundamental identity

#sin^2(x)+cos^2(x)=1#