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Why does $$-\frac{\cos^2 x}{\sin x}= -\cos x\cot x?$$ Sorry for the dumb question.

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  • $\begingroup$ break up the $-cos^2(x)$ into a product of 2 cosines. then use a trig definition. $\endgroup$ – MathHype Mar 28 '15 at 19:29
  • $\begingroup$ Remember that $a^2 = aa$, and that $\frac{ab}{c} = \frac{a}{c}b = a\frac{b}{c}$. $\endgroup$ – Miguelgondu Mar 28 '15 at 19:40
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Look at my comment above. $$\frac{-\cos^2(x)}{\sin(x)}\rightarrow \frac{-\cos(x)*{\color{red} {\cos(x)}}}{\color{red}{\sin(x)}} \rightarrow -\cos(x)\cot(x)$$

hint : $\cot(x)=\frac{cos(x)}{\sin(x)}$

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  • $\begingroup$ Forgot a minus in the hint, :) $\endgroup$ – Miguelgondu Mar 28 '15 at 19:35
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    $\begingroup$ @Miguelgondu Hah, i was using that as a separator. I can see how that can me miss understood. I will replace with a colon, thanks! $\endgroup$ – MathHype Mar 28 '15 at 19:37
  • $\begingroup$ Oh wow it makes sense now! I should've separated into two cosines facepalm $\endgroup$ – user3709169 Mar 28 '15 at 19:42
  • $\begingroup$ @user3709169 if you like the red for emphasis, check mine as accepted! Also, no need to facepalm, my first idea was wrong, this was my second attempt. Just something to get used to with math. Stay persistent! $\endgroup$ – MathHype Mar 28 '15 at 19:44
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$-\frac{\cos^2(x)}{\sin(x)}=-\frac{\cos(x)\cdot \cos(x)}{\sin(x)}=-\cos(x)\cdot\frac{\cos(x)}{\sin(x)}=-\cos(x)\cot(x)$

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