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Simplify: $$\sin(x-2\pi)\cos\left(\frac{3\pi}{2}-x\right)+\tan(\pi-x)\tan\left(\frac{3\pi}{2}+x\right)$$

What I did was convert each term to its simplest form.

$$\sin(x-2\pi)\cos\left(\frac{3\pi}{2}-x\right) = \sin(x)\sin(x) = \sin^2(x)$$

$$\tan(\pi-x)\tan\left(\frac{3\pi}{2}+x\right) = (-\tan(x))(-\cot(x))=1 $$

So than I added those two together and got:

$$\sin^2(x)+1$$

However this is wrong somehow. Any ideas?

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  • $\begingroup$ $\cos((3\pi/2)-x) = \cos(3\pi/2)\cos(x) + \sin(3\pi/2)\sin(x) = 0\cos(x) + (-1)\sin(x) = -\sin x$. I think you also have the wrong sign for $\tan((3\pi/2) + x)$... $\endgroup$ – Arturo Magidin Sep 30 '18 at 0:10
  • $\begingroup$ Arturo, we still haven't "learned" the sum and difference identities, so how would i solve otherwise. $\endgroup$ – Ozymandias Sep 30 '18 at 0:32
  • $\begingroup$ $\cos(x)$ is even; so $\cos((3\pi/2)-x)$ is the same as $\cos(x-(3\pi/2))$. Now, the graph of $f(x-a)$ is the graph of $f$ translated right by $a$. The graph of $\cos(x)$, translated to the right by $3\pi/2$ Is the graph of $-\sin(x)$, not of $\sin(x)$. Similarly, the graph of $\cos(x+(3\pi/2))$ is the graph of $\cos(x)$ translated left by $3\pi/2$, which is the graph of $\sin(x)$, and the graph of $\sin((3\pi/2)+x)$ is the graph of $\sin(x)$ translated left by $3\pi/2$, which is the graph of $-\cos(x)$. (Cont) $\endgroup$ – Arturo Magidin Sep 30 '18 at 0:38
  • $\begingroup$ Alright :) Thank you $\endgroup$ – Ozymandias Sep 30 '18 at 0:39
  • $\begingroup$ So $\tan((3\pi/2)+x) = \sin((3\pi/2)+x)/\cos((3\pi/2)+x) = -\cos(x)/sin(x) = -\cot(x)$ (maybe that sign was right). The graph of $\tan(\pi-x)$ is the same as the graph of $\tan(-x)$ by periodicity, and $\tan(-x)=-\tan(x)$. $\endgroup$ – Arturo Magidin Sep 30 '18 at 0:40
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Hint: $$\begin{split} &\sin(x-2\pi)\cos(3\pi/2-x)+\tan(\pi-x)\tan(3\pi/2+x)\\ = &-\sin(x)\cos(\pi/2-x) + \left(-\tan(x)\right)\left(-\tan(x-\pi/2)\right)\\ =& -\sin^2(x) + \tan(x)\tan(x-\pi/2) \end{split}$$

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