Algebranometry

Geometry Level 3

If ( 1 + 1 + t ) tan ( t ) = 1 + 1 t \left(1+\sqrt { 1+t } \right)\tan(t)=1+\sqrt { 1-t } for real t t and f = log t sin ( 4 t ) f=\log _{ t }{ \sin(4t) } , find ( f + 1 ) ( f 2 + 1 ) ( f 3 + 1 ) (f+1)({ f }^{ 2 }+1)({ f }^{ 3 }+1) .


The answer is 8.

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1 solution

Mark Hennings
Nov 27, 2018

We have (after some simplification) tan t = 1 + 1 t 1 + 1 + t tan 2 t = 2 tan t 1 tan 2 t = 1 + 1 t 2 t sin 4 t = 2 tan 2 t 1 + tan 2 2 t = t \begin{aligned} \tan t & = \; \frac{1 + \sqrt{1-t}}{1 + \sqrt{1+t}} \\ \tan2t & = \; \frac{2\tan t}{1 - \tan^2t} \; = \; \frac{1 + \sqrt{1-t^2}}{t} \\ \sin4t & = \; \frac{2\tan2t}{1 + \tan^22t} \; = \; t \end{aligned} so that f = 1 f=1 , making the answer 8 \boxed{8} .

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