Trigonometry
20 Online
OpenStudy (danielajohana):
Rewrite sin(x)cos^2 (x)-sin(x) as an expression containing a single term.
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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.
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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
\[ab^2 - a~:~factor,~-a\]
\[-a(1-b^2)\]
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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?
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OpenStudy (amistre64):
there is no +2, 1-1 = 0 not 2