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Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out.

$$ \begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-x\right) +\cot(2\pi+x) \right] \\ &= \sin x\cos x[\tan x+\cot x]\end{split} $$ I've tried the following: $\cot(3\pi/2)$ is $0$, so it would leave out $-\cot x$, that's where I lose the thread. $\cos(3\pi/2)$ is 0, so that leaves out $\cos x$ and $\cos2\pi$ is one and that's where I lose the thread.

Note: that this is not an assignment question, these are all solved examples, only for my practice. Source image

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    $\begingroup$ Just so you know, and perhaps make sense of the answers, $\cos(a \pm b) = \cos a \cos b \mp \sin a\sin b$. $\endgroup$
    – amWhy
    Aug 15, 2014 at 15:28
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    $\begingroup$ See math notation guide. $\endgroup$
    – user147263
    Aug 15, 2014 at 17:49

3 Answers 3

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Using the formula $$\cos{(a+b)}=\cos{(a)} \cos {(b)}-\sin{(a)} \sin{(b)}$$ we get the following:

$$\cos{ \left ( \frac{3 \pi}{2}+x \right )}=\cos{ \left ( \frac{3 \pi}{2}\right )} \cos{(x)}-\sin{ \left ( \frac{3 \pi}{2} \right )} \sin{(x)}=0 \cdot \cos{(x)}-(-1) \cdot \sin{(x)}=\sin{(x)}$$

$$\cos{(2 \pi+x)}=\cos{(2 \pi)} \cos{(x)}-\sin{(2 \pi )} \sin{(x)}=\cos{(x)}$$

Use the formula $$\cot{(a+b)}=\frac{\cos{(a)} \cos{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}} \\ \text{ and } \\ \cot{(a-b)}=-\frac{\cos{(a)} \cos{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}$$ to calculate $\cot{\left (\frac{3\pi}{2}-x \right )}$ and $\cot{(2 \pi+x)}$.

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The $\cos(2\pi+x)$ and $\cot(2\pi+x)$ terms are easy to get rid of: the basic trigonometric functions have period $2\pi$. So $\cos(2\pi +x)=\cos x$, and $\cot(2\pi+x)=\cot x$.

As to $\cos(3\pi/2+x)$, we can use the addition formula. It is $\cos(3\pi/2)\cos x-\sin(3\pi/2)\sin x$. This is simply $\sin x$. That's because $\cos(3\pi/2)=0$ and $\sin(3\pi/2)=-1$.

Your turn. You need to deal with $\cot(3\pi/2 -x)$. Use the procedure of the preceding paragraph, using the fact that $\cot t=\frac{\cos t}{\sin t}$.

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  • $\begingroup$ Thank you! Because I added explanatory material, your name got erased. But probably there is at least one more typo to fix. $\endgroup$ Aug 15, 2014 at 15:26
  • $\begingroup$ I can certainly relate regarding typos! :-| $\endgroup$
    – amWhy
    Aug 15, 2014 at 15:29
  • $\begingroup$ By periodicity, $\cos(2\pi-x)=\cos(2\pi+(-x))=\cos(-x)$. But $\cos(-x)=\cos x$ for all $x$, so the book is right. $\endgroup$ Aug 15, 2014 at 15:31
  • $\begingroup$ How is cos(2π+x) = cox? It is given in my book that cos(2π-x)= cosx, is cos(2π+x) = cosx as well? $\endgroup$ Aug 15, 2014 at 15:32
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    $\begingroup$ Yes. For $\sin$ and $\cos$, and therefore also for all the other trig fumctions, we have periodicity. In general $\cos(x+2n\pi)=\cos(x)$ for any integer $n$. Same for $\sin(x+2n\pi)$. Looking at pictures of $\cos$ and $\sin$ can be very useful. Shift the curve by $2\pi$, and the picture does not change. $\endgroup$ Aug 15, 2014 at 15:36
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Replace the terms on the left side of your left side of the identity with these, and you will end up with the right hand side of your identity.

$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$

$$\require{cancel}\cos \left(2\pi+x \right)=\cancelto{1}{\cos 2\pi}\cos x - \cancelto{0}{\sin 2 \pi}\sin x=\cos x$$

$$\require{cancel} \cot\left(\frac{3\pi}2 - x\right) = \frac{\cos(\frac{3\pi}2 - x)}{\sin(\frac{3\pi}2 - x)} = \frac{\cancelto{0}{\cos\frac{3\pi}2} \cos x + \cancelto{1}{\sin\frac{3\pi}2} \sin x }{\cancelto{1}{\sin\frac{3\pi}2} \cos x - \cancelto{0}{\cos\frac{3\pi}2} \sin x} = \frac{\sin x}{\cos x} = \tan x$$

$$\require{cancel} \cot\left(2\pi + x\right) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cancelto{1}{\cos 2\pi} \cos x - \cancelto{0}{\sin 2\pi} \sin x }{\cancelto{0}{\sin 2\pi} \cos x + \cancelto{1}{\cos 2\pi} \sin x} = \frac{\cos x}{\sin x} = \cot x$$

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  • $\begingroup$ How did you get those two last terms in both of the results? $\endgroup$ Aug 15, 2014 at 15:24
  • $\begingroup$ Not those last two terms. In the first two equations YOU wrote, how did you get sin3pi/2 and sinx? $\endgroup$ Aug 15, 2014 at 15:27
  • $\begingroup$ Use the formula: $$\cos(a+b)=\cos a \cos b - \sin a \sin b$$ $\endgroup$
    – Cookie
    Aug 15, 2014 at 15:27
  • $\begingroup$ Okay...fine, fine fine. thanks $\endgroup$ Aug 15, 2014 at 15:28

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