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Trigonometry 20 Online
OpenStudy (danielajohana):

Rewrite sin(x)cos^2 (x)-sin(x) as an expression containing a single term.

OpenStudy (amistre64):

factor, negate, identity, multiply

OpenStudy (amistre64):

if we factor and negate as one step then thats fine too

OpenStudy (danielajohana):

can you explain it using the equation please?

OpenStudy (saifoo.khan):

Take commons for factor.

OpenStudy (danielajohana):

I'm sorry; I'm visual, so I need to see how to do it, if that's possible.

OpenStudy (saifoo.khan):

sinx?

OpenStudy (amistre64):

you have\[ab^2-a\]factor out the -a

OpenStudy (danielajohana):

we factor the sinx?

OpenStudy (amistre64):

-sin(x) yes

OpenStudy (amistre64):

it may be simpler to follow it with less clutter tho \[ab^2 - a~:~factor,~-a\] \[-a(1-b^2)\]

OpenStudy (amistre64):

now in terms of the trig used, 1-b^2 has an identity in terms of a

OpenStudy (danielajohana):

So I would replace cosx with its sin equivalent?

OpenStudy (saifoo.khan):

Yes.

OpenStudy (danielajohana):

cos^2 (x)= -sin^2(x)+1 so: -sinx(1-(-sin^2 (x)+1)) -sinx(1+sin^2 (x)+1) -sinx(sin^2(x)+2) now what do I do?

OpenStudy (danielajohana):

wait, is that the answer?

OpenStudy (amistre64):

there is no +2, 1-1 = 0 not 2

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