I curious practical solution.(Step by step)
$\left( 1+\tan 5^\circ\right) \left( 1+\tan 10^\circ\right)\left( 1+\tan 15^\circ\right)\cdots\left( 1+\tan 40^\circ\right)$
Answer is $16$.
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Do you mean degree instead of radian? – Mar 28 '16 at 08:28
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@frank000 sorry.degree – FMath Mar 28 '16 at 08:30
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Did you come up with any idea? – jonsno Mar 28 '16 at 08:30
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What do you mean by "this sum" ?? – Mar 28 '16 at 08:31
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Isn't it a product? – Kaster Mar 28 '16 at 08:32
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Related : http://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt – lab bhattacharjee Mar 28 '16 at 08:47
2 Answers
11
If $\displaystyle A+B = \frac{\pi}{4}\;,$ Then $(1+\tan A)(1+\tan B) = 2$
So $(1+\tan 5^0)(1+\tan 40^0)=2$ and $(1+\tan 10^0)(1+\tan 35^0)=2$
and $(1+\tan 15^0)(1+\tan 30^0)=2$ and $(1+\tan 20^0)(1+\tan 35^0)=2$
So $$(1+\tan 5^0)(1+\tan 40^0)(1+\tan 10^0)(1+\tan 35^0)(1+\tan 15^0)(1+\tan 30^0)(1+\tan 20^0)(1+\tan 35^0)=2^4=16$$
juantheron
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For example, observe that $$1=\tan 45^\circ =\tan(5^\circ+40^\circ)=\frac{\tan5^\circ+\tan40^\circ}{1-\tan5^\circ\tan40^\circ}.$$ Then we have $\tan5^\circ+\tan40^\circ=1-\tan5^\circ\tan40^\circ$, and thus \begin{align} (1+\tan5^\circ)(1+\tan40^\circ)&=1+\tan5^\circ\tan40^\circ+\tan5^\circ+\tan40^\circ\\ &=1+\tan5^\circ\tan40^\circ+1-\tan5^\circ\tan40^\circ\\ &=2. \end{align} Hence we conclude that $$(1+\tan5^\circ)(1+\tan10^\circ)\cdots(1+\tan40^\circ)=2^4=16.$$
Solumilkyu
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