Apparently,
$$(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000000000000\cdots$$
Since it is a $2.0000000000\cdots$ instead of $2$, it isn't exactly $2$.
Why is that?
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Aditya Hase
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Kurt
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7The RHS of the equality is exactly $2$. – Git Gud Nov 27 '14 at 01:30
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5Hint: $37^\circ + 8^\circ = 45^\circ$. – MJD Nov 27 '14 at 01:37
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1Not sure how to answer... perhaps you could explain why you believe $2.0000\cdots$ is not the same as $2$? – David Nov 27 '14 at 01:37
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6Use the addition law for $\cot(x+y)$, or the more familiar $\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$. – André Nicolas Nov 27 '14 at 01:38
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OK, I hope my answer below is simpler than some I see here. – Michael Hardy Nov 27 '14 at 02:49
5 Answers
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Using various trigonometric identities, you should be able to show the left hand side is equal to $2$. The ones you will need are: $$\cos(A+B)=\cos A\cos B-\sin A\sin B,\quad \cos(-A)=\cos A,\quad \cos(45^\circ)=\sin(45^\circ)=\frac1{\sqrt{2}}$$ as well as an understanding of what $\cot A$ means. The reason you have the answer of $2.000000\ldots$ (which I assume you got from WolfRamAlpha or similar) is because it's easier for a computer to just work out $\cot 37^\circ,\cot 8^\circ$ to any degree of accuracy and then solve by brute force. So the computer gets an answer of, for example, $2.000000$ correct to $6$ decimal places, but cannot be sure that the answer is exactly $2$.
Jason
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You don't need to write $37^o$; you can write $37^\circ$. (I changed it.) ${}\qquad{}$ – Michael Hardy Nov 27 '14 at 01:52
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$$\begin{align} 1-\cot{\left(x\right)} &=1-\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}}\\ &=\frac{\sin{\left(x\right)}-\cos{\left(x\right)}}{\sin{\left(x\right)}}\\ &=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}.\\ \end{align}$$
Therefore,
$$\begin{align} 1-\cot{\left(\frac{\pi}{4}-x\right)} &=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-x\right)\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}\\ &=-\sqrt{2}\frac{\sin{\left(x\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}.\\ \end{align}$$
Hence,
$$\begin{align} \left(1-\cot{\left(x\right)}\right)\cdot\left(1-\cot{\left(\frac{\pi}{4}-x\right)}\right) &=\left[-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}\right]\cdot\left[-\sqrt{2}\frac{\sin{\left(x\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}\right]\\ &=2.\ \end{align}$$
David H
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$$ 1 = \cot45^\circ = \cot(37^\circ+8^\circ) = \frac{\cot37^\circ\cot8^\circ-1}{\cot37^\circ+\cot8^\circ}. $$ Therefore $$ \cot37^\circ+\cot8^\circ = \cot37^\circ\cot8^\circ-1 $$ so $$ 2 = 1-\cot37^\circ-\cot8^\circ +\cot37^\circ\cot8^\circ=(1-\cot37^\circ)(1-\cot8^\circ). $$
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Alternatively, $$\cot(x+y) = \frac{\cos(x+y)}{\sin(x+y)} = \frac{\cos x \cos y - \sin x \sin y}{\sin x \cos y + \cos x \sin y} = \frac{\cot x \cot y - 1}{\cot x + \cot y}$$
Now if $x = 37^\circ$, $y = 8^\circ$, then $\cot(x + y) = \cot 45^\circ = 1$ and
$$\begin{align} \ \ \ \ 1 & = \frac{\cot x \cot y - 1}{\cot x + \cot y} \\ \cot x + \cot y & = \cot x \cot y - 1 \\ 2 + \cot x + \cot y & = \cot x \cot y + 1 \\ 1 - \cot x - \cot y + \cot x \cot y & = 2 \\ (1 - \cot x)(1 - \cot y) & = 2 \\ (1 - \cot 37^\circ)(1 - \cot 8^\circ) & = 2 \end{align}$$
Simon S
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$2.\dot 0$ is exacyly equal to 2
Paul
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But how do you know it keeps repeating forever, as opposed to $(1-\cot37^\circ)(1-\cot8^\circ)$ being $2.00000\ldots00003$, with the "$3$" in the ten-thousandth place? ${}\qquad{}$ – Michael Hardy Nov 27 '14 at 01:53
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