3
$\begingroup$

What is the strategy to solve for:

$$(1+\tan1^\circ)(1+\tan2^\circ)(1+\tan3^\circ)\cdots(1+\tan45^\circ)$$

I don't know where to start. Thanks.

  • 0
    That looks like a product, not a sum. – 2017-02-07
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    This is a product, not a sum, FYI. – 2017-02-07
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    Great minds, eh, @arctictern? – 2017-02-07
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    Try and pair up opposing terms with the angles summing to $45$. Use any formula you know for $\tan$ to help you (Hint: Can you say that $(1+ \tan 1)(1 + \tan 44) = 2$?) – 2017-02-07
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    Are you sure the angles are in radians and not in degrees? – 2017-02-07
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    Angles are in degrees – 2017-02-07
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    "Solve" seems to be one of those catch-all terms for people who don't know what word to use. "Evaluate" would be appropriate here. One solves problems; one solves equations; one _evaluates_ expressions. – 2017-02-07
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    Related : http://math.stackexchange.com/questions/1716859/how-to-calculate-left-1-tan-5-circ-right-left-1-tan-10-circ-right-lef and http://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt – 2017-02-07

1 Answers 1

7

Below is the answer. You will need to apply trigonometric identities:

$$\tan(\alpha + \beta) = \frac{ \tan(\alpha) + \tan(\beta) } { 1 - \tan(\alpha) \tan(\beta)}$$

Note that: $$\tan(45^\circ) = 1$$

Thus, $$\tan(45^\circ) = \frac{\tan1^\circ + \tan44^\circ }{ 1 - \tan1^\circ \tan44^{\circ}}$$

$$\tan(45^\circ) = \frac{\tan2^\circ + \tan43^\circ }{1 - \tan2^\circ \tan43^\circ}$$

$$\vdots$$

$$\tan(45^{\circ}) = \frac{\tan22 + \tan23}{1 - \tan22 \tan23}$$

Then, it follows:

$$\tan1^{\circ} + \tan44^{\circ} = \tan45^{\circ} (1 - \tan1^{\circ} \tan44^{\circ})$$

$$\tan2^{\circ} + \tan43^{\circ} = \tan45^{\circ} (1 - \tan2^{\circ} \tan43^{\circ})$$

$$\vdots$$

$$\tan22^{\circ} + \tan23^{\circ} = \tan45^{\circ}(1 - \tan22^{\circ} \tan23^{\circ})$$

\begin{align*} & (1 + \tan1) (1 + \tan2) (1 + \tan3) \cdots(1 + \tan44) (1 + \tan45) \\[10pt] = {} & (1 + \tan1) (1 + \tan2) (1 + \tan3) \cdots(1 + \tan44) (1 + 1) \\[10pt] = {} & 2 (1 + \tan1) (1 + \tan2) (1 + \tan3) \cdots (1 + \tan44) \\[10pt] = {} & 2 (1 + \tan1) (1 + \tan44) (1 + \tan2) (1 + \tan43)(1 + \tan3) (1 + \tan42) \cdots (1 + \tan22) (1 + \tan23) \\[10pt] = {} & 2 (\tan1 + \tan44 + \tan1 \tan44 + 1) (\tan2 + \tan43 + \tan2 \tan43 + 1) \cdots (\tan22 + \tan23 + \tan22 \tan23 + 1) \\[10pt] = {} & 2 ((\tan45)(1 - \tan1 \tan44) + \tan1 \tan44 + 1) ((\tan45)(1 - \tan2 \tan43) + \tan2 \tan43 + 1) \cdots ((\tan45)(1 - \tan22 \tan23) + \tan22 \tan23 + 1) \\[10pt] = {} & 2 (1 - \tan1 \tan44 + \tan1 \tan44 + 1) (1 - \tan2 \tan43 + \tan2 \tan43 + 1) \cdots (1 - \tan22 \tan23 + \tan22 \tan23 + 1) \\[10pt] = {} & 2 \times \cdots \times 2 \\[10pt] = {} & 2 \times (2^{22}) = 2^{23}\\[10pt] = {} & \boxed{8388608} \end{align*}

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    There is no way this is what the OP was asking for. – 2017-02-07
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    @TheCount Actually, I think it is. – 2017-02-07
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    This needs a lot of Mathjax help. First step - replace `tan` everywhere with `\tan`. And the line breaks are terrifying. Also `2^{22}`, not `2^22`, which formats as $2^22$. – 2017-02-07
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    How about just noting that $\tan a +\tan(45-a)=1-\tan a\tan(45-a)$. No reason to write out all those lines. Way too much noise in this when the idea is nice and elegant. – 2017-02-07
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    Fixed formatting issue in my Latex text editor instead of the web. – 2017-02-07
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    @arctictern it looks a lot more reasonable now than when i made the comment, to be fair – 2017-02-07
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    Answer looks fine, ut how do i know that i have to do product this way. Is this a general method used to find product of trig series – 2017-02-07
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    The "align" section is still terrible. Just say $(1+\tan x)(1+\tan(45-x))=...=2$ then put it together. – 2017-02-07
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    1 radian is not the same as 1 degree. – 2017-02-07