Trigonometry II

Geometry Level 3

α = ( 3 + tan ⁡ ( 1 ∘ ) ) . ( ( 3 + tan ⁡ ( 2 ∘ ) ) . . . ( ( 3 + tan ⁡ ( 2 9 ∘ ) ) \displaystyle \alpha = (\sqrt{3} + \tan (1^\circ)).((\sqrt{3} +\tan(2^\circ))...((\sqrt{3}+\tan(29^\circ))

What Is The Last Three Digits of α \alpha ?

Details :

  • Is Product Tan 1 ∘ 1^\circ Until 2 9 ∘ 29^\circ

  • You Can Use A Calculator


The answer is 912.

Gabriel Merces by Gabriel Merces , 23, Brazil

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1 solution

Dinesh Chavan
Apr 4, 2014

It is given that α = ( 3 + t a n ( 1 ) ) . ( 3 + t a n ( 2 ) ) . ( 3 + t a n ( 3 ) ) . . . . . . ( 3 + t a n ( 29 ) ) \alpha=(\sqrt{3}+tan(1)).(\sqrt{3}+tan(2)).(\sqrt{3}+tan(3))......(\sqrt{3}+tan(29)) Which can be written as α = ( 3 + t a n ( 1 ) ) . ( 3 + t a n ( 29 ) ) . ( 3 + t a n ( 2 ) ) . ( 3 + t a n ( 28 ) ) . . . . . . . ( 3 + t a n ( 14 ) ) . ( 3 + t a n ( 16 ) ) . ( 3 + t a n ( 15 ) ) \alpha=(\sqrt{3}+tan(1)).(\sqrt{3}+tan(29)).(\sqrt{3}+tan(2)).(\sqrt{3}+tan(28)).......(\sqrt{3}+tan(14)).(\sqrt{3}+tan(16)).(\sqrt{3}+tan(15))

Now,, notice that ( 3 + t a n ( 1 ) ) . ( 3 + t a n ( 29 ) ) = 3 + 3 ( t a n ( 1 ) + t a n ( 29 ) ) + t a n ( 1 ) . t a n ( 29 ) . . . . . . . . . . . ( 1 ) (\sqrt{3}+tan(1)).(\sqrt{3}+tan(29))=3+\sqrt{3}(tan(1)+tan(29))+tan(1).tan(29)...........(1)

Now we know from the formula that t a n ( 30 ) = t a n ( 1 ) + t a n ( 29 ) 1 − t a n ( 1 ) t a n ( 29 ) tan(30)=\frac{tan(1)+tan(29)}{1-tan(1)tan(29)}
Therefore, t a n ( 1 ) + t a n ( 29 ) = 1 3 ( 1 − t a n ( 1 ) t a n ( 29 ) ) tan(1)+tan(29)=\frac{1}{\sqrt{3}}(1-tan(1)tan(29)) , so putting this value in ( 1 ) (1) , we get ( 3 + t a n ( 1 ) ) . ( 3 + t a n ( 29 ) ) = 3 + ( 1 − t a n ( 1 ) . t a n ( 29 ) ) + t a n ( 1 ) . t a n ( 29 ) = 4 (\sqrt{3}+tan(1)).(\sqrt{3}+tan(29))=3+(1-tan(1).tan(29))+tan(1).tan(29)=4

So, like this, all the pairs cancel out to give 4 4 in each bracket, So, the original problem becomes α = 4 14 ( ( 3 + t a n ( 15 ) ) \alpha=4^{14}((\sqrt{3}+tan(15))

Now, we see that ( 3 + t a n ( 15 ) ) = 3 + s i n ( 30 ) 1 + c o s ( 30 ) = 2 (\sqrt{3}+tan(15))=\sqrt{3}+\frac{sin(30)}{1+cos(30)}=2

So, the original number α \alpha is nothing but, α = ( 4 14 ) . 2 \alpha=(4^{14}).2

Now, we may either compute 4 14 4^{14} , using a standard calculator, or we may cork it with congruence to obtain the last 3 digits as 456 456 Now, Last three digits of α = 456 × 2 = 912 \alpha=456×2=912

8 Helpful 0 Interesting 0 Brilliant 0 Confused

in step compute 4^14 how can you find last three digit?

Jo Sg - 6 years, 8 months ago

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