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How do I simplify tan$\theta$ cos$\theta$ ?

Why is this so hard to do? What pieces of information should I know before doing these?

Can someone just tell me were am I going wrong? I have 5 days to master this before my SAT Practice.

I would brush up on 'basics' but i don't even know how this related to anything ive learnt in trig before...I literally google every identity, am i supposed to remember or somehow derive them??

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    $\begingroup$ Do you remember the definition of tan? $\endgroup$ – Mathmo123 Jul 22 '14 at 10:27
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    $\begingroup$ Do you know how tan relates to sin and cos? (4 days, 23h, 58min) $\endgroup$ – Lord_Gestalter Jul 22 '14 at 10:28
  • $\begingroup$ sosmath.com/trig/Trig5/trig5/trig5.html $\endgroup$ – Scientifica Jul 22 '14 at 10:28
  • $\begingroup$ To use the 5 days brushing up the basics will be worth more than collecting and memorising possible answers $\endgroup$ – Lord_Gestalter Jul 22 '14 at 10:34
  • $\begingroup$ I saw your earlier posts. You are going wrong, by not concentrating on the basic definition of sine, cos and tan. If I was you, I would take a pen and paper and start by drawing triangle and learn the ratios of sin,cos and tan. It would not be called maths, if it was not tricky! :) $\endgroup$ – MonK Jul 22 '14 at 10:35
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Hint: What is the definition of the $\tan$ function?

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  • $\begingroup$ Tan = Opp / Adj.. $\endgroup$ – sasha Jul 22 '14 at 10:38
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    $\begingroup$ @sasha That is a rough definition only true for angles from $0$ to $\frac\pi2$. For other corners, the $\tan$ is defined via the $\cos$ and $\sin$ function. If you are practising for SATs, I advise you to look that definition up. $\endgroup$ – 5xum Jul 22 '14 at 10:40
  • $\begingroup$ @sasha $\tan\theta = opp/adj = \displaystyle\frac{opp}{adj}\cdot\frac{\frac{1}{hyp}}{\frac{1}{hyp}} = \frac{\frac{opp}{hyp}}{\frac{opp}{hyp}}=\frac{\sin\theta}{\cos\theta}$ $\endgroup$ – John Joy Jul 22 '14 at 13:30
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Notice first that the $\tan$ function is defined on $D:=\Bbb R\setminus\{\frac\pi2+k\pi,\; k\in\Bbb Z\}$ and that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ so

$$\tan\theta\cos\theta=\sin\theta,\quad\forall\theta\in D$$ and be careful the trap is to give a wrong domain.

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  • $\begingroup$ I am used to write - rather than $\$ for excluding. Anyway +1 $\endgroup$ – mrs Jul 22 '14 at 12:21
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Elementary school proof for $\theta \in ]0, \pi/2[$:

Consider a right triangle with angles $\pi/2, \theta, \pi/2-\theta$ and edges lengths $a,b,c>0$ as in the figure below

Right triangle

Then $$\tan(\theta)= \frac{a}{c}, \quad \cos(\theta) = \frac{c}{b}, \quad \sin(\theta) = \frac{a}{b},$$ thus $$\tan(\theta) \cos(\theta) = \frac{a}{c}\cdot \frac{c}{b} = \frac{a}{b} = \sin(\theta).$$

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There are two ways to do this. Here is one way:

Saying that

$\tan(x)=o/a $ and $ \cos(x)=a/h$, we get $oa/ha = o/h=\sin x$ so the answer is $\sin x$!

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