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SugarRainbow
combine the fractions and simplify to a multiple of a power of a basic trigonometric function sinx/cot^2x-sinx/cos^2x?
can i change the sinx/cos^2x into tan^2x?
Hmm no. You can take ONE of the cosines, and change it to tanx*(1/cosx) But I dont know if that's the right way to go c: hmm
\[\large \frac{ \sin x }{ \cot^2 x }-\frac{ \sin x }{ \cos^2 x }=\frac{ \sin x }{ (\frac{ \cos^2 x }{ \sin^2 x }) }-\frac{ \sin x }{ \cos^2 x }\] \[\large \frac{ \sin x }{ 1 }*\frac{ \sin^2 x }{ \cos^2 x }-\frac{ \sin x }{ \cos^2 x }\]
There are multiple ways to simplify it, since there are so many trig identities, so you might find a different way... but this is what I was thinking :3
\[\large \frac{ \sin^3 x - \sin x }{ \cos^2 x }\]
Confused about any of that so far? :D
how did you get sinx/1∗sin^2x/cos^2x
\[\large \frac{ \left(\frac{ a }{ b }\right) }{ \left(\frac{ c }{ d }\right) }=\frac{ a }{ b }*\frac{ d }{ c }\]
Remember how you can do that? when dividing fractions :D flip the bottom one and rewrite it as multiplication.
but the top wasn't a fraction yet it was sinx/ (cos^2x/sin^2x)
oh wait you put it over 1 to make it into a fraction huh?
\[\huge \frac{ \left(\sin x \right) }{ \left(\frac{ \cos^2 x }{ \sin^2 x } \right) }=\frac{ \left(\frac{ \sin x }{ 1 } \right) }{ \left(\frac{ \cos^2 x }{ \sin^2 x } \right) }\]
I gotta go, so lemme post the rest of the steps, and you can at least look at them while I'm gone. I won't be here to answer questions though :c
\[\large \frac{ \sin^3 x - \sin x }{ \cos^2 x }=\frac{ \sin x(\sin^2 x - 1) }{ 1-\sin^2 x }\]
Factored a sinx from each term in the top. Rewrote cos^2 as 1-sin^2 using identity :D
Now if we factor a -1 ouch of each term in the top brackets, we'll have a nice easy cancellation.
\[\large \frac{ -\sin x(1-\sin^2 x) }{ 1-\sin^2 x }=-\sin x\]
Hopefully I didn't make a mistake in there :D heh
alright well thanks