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Is it true that $\tan x\cdot \cos x = \sin x$? If I put $x=30$ in my calculator then I don't get the same answer as $\sin 30$, why is this? Don't the two cosines cancel out? I'm probably missing something really stupid here.

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    $\begingroup$ Have you inputted $\tan(30^\circ)\cos(30^\circ)$ or something that will give you $\tan(30\cos(30))$? Also this is not true for $x=(90^\circ$ plus multiples of $180^\circ$). $\endgroup$ – JP McCarthy Apr 27 '15 at 10:42
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    $\begingroup$ You might want to double-check that your calculator is on degrees. They both give sin(30), but only with degrees do you have sin(30)=1/2. $\endgroup$ – Akiva Weinberger Apr 27 '15 at 10:47
  • $\begingroup$ (But tanx*cosx=sinx, yeah. Assuming tanx is defined.) $\endgroup$ – Akiva Weinberger Apr 27 '15 at 10:47
  • $\begingroup$ It is possible only when $\cos x$ is non-zero $\endgroup$ – Empty Apr 27 '15 at 10:48
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    $\begingroup$ A properly functioning calculator used properly will return the same value for $\tan 30\cdot\cos 30$ as for $\sin 30$ regardless of whether it is set on degrees or radians or any other unit, except when the unit is so chosen that the number $30$ represents a right angle or other value for which the cosine is $0$. However, I said "used properly" and I've noticed students sometimes getting really confused about that and entering things like $\tan(30\sin30)$ instead of $\tan30\cdot\sin30$, etc. Maybe some explicit report on the sequence of key strokes could shed some light on what's going on. $\endgroup$ – Michael Hardy Apr 27 '15 at 11:40
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By definition, $$ \tan x=\frac{\sin x}{\cos x} $$ but in order to give sense to this definition, the $\cos$ must be nonzero, i.e. $x\neq\frac{\pi}2+k\pi$. Hence $\tan$ is defined only for $x\in\Bbb R\setminus \{\frac{\pi}{2}+k\pi\}$. For these values you can multiply both sides for $\cos$, obtaining $\tan x\cos x=\sin x$.

But pay attention: this holds only for the values I wrote! Otherwise $\tan$ is not even defined, so it wouldn't make sense to consider the relation.

Now in your case, $x=30°$ is the same as $x=\pi/6$ in radiant and in this value the $\tan$ is defined. So the relation holds.

You must had a bad work with your calculator, that's all.

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As tan(x)≡ Sin(x)/Cos(x), you are right in that Tan(x) * cos(x) ≡ Sin(x). This is by definition of the Tan function, which is defined as Sin(x) / Cos(x). If it helps consider the right angle triangle from the unit circle, where cos(x) = Hypotenuse / adjacent and Sin(x) = opposite / hypotenuse, so Tan(x) as equalling opposite / adjacent is easy to see by dividing one by the other since the hypotenuse will cancel in both. However, it is worth remembering that for certain values, where x is either an odd multiple of 90 degrees or odd multiples of Pi/2 if dealing in radians, this simply does not work as Cos(x) = 0 ( it is easy enough to see that at 90 degrees, this leads to 1 = 0 which is obviously nonsensical).

However, it is easy enough to prove that since Sin(30) = 0.5 and Cos(30) = root(3) /2, then tan(30) = 1 / root(3). Multiplying this by cos(30) does in fact lead us to 0.5 again, so this is right here (luckily the normal rules of communativity apply here.) However, I can see how with approximation in the calculator it would not work. If you try using the exact answers this should be proven to be true.

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Without doubt,

$$\tan30°\cdot\cos30°=\frac1{\sqrt3}\cdot\frac{\sqrt3}2=\frac12=\sin30°$$ $$0.57735026919\cdots\times0.86602540378\cdots=0.5$$

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