Tangents of nonstandard angles?

Geometry Level 4

3 ( 1 3 + tan ⁡ 1 ∘ ) ( 1 3 + tan ⁡ 2 ∘ ) ⋯ ( 1 3 + tan ⁡ 5 8 ∘ ) ( 1 3 + tan ⁡ 5 9 ∘ ) \sqrt3 \left(\frac{1}{\sqrt3}+\tan1^\circ\right)\left(\frac{1}{\sqrt3}+\tan2^\circ\right)\cdots\left(\frac{1}{\sqrt3}+\tan58^\circ\right)\left(\frac{1}{\sqrt3}+\tan59^\circ\right)

The above product can be written as a b c d \dfrac{a^{b}}{c^{d}} , where a , c a, c are prime numbers. Then find a + b + c + d a+b+c+d upto 2 decimal places.

Bonus :

Try to generalize (I think you'll get what I'm saying once you solve the problem).

This is problem is a more modified form of a question from our tests.
Feel this is kinda simple, try this problem


The answer is 93.00.

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Vignesh S by Vignesh S , 22, India

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4 solutions

Sam Bealing
Apr 15, 2016

Use the identity:

tan ⁡ A + tan ⁡ B = sin ⁡ A cos ⁡ A + sin ⁡ B cos ⁡ B = sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A cos ⁡ A cos ⁡ B = sin ⁡ ( A + B ) cos ⁡ A cos ⁡ B \tan{A}+\tan{B}=\frac{\sin{A}}{\cos{A}}+\frac{\sin{B}}{\cos{B}}=\frac{\sin{A} \cos{B}+\sin{B} \cos{A}}{\cos{A} \cos{B}}=\frac{\sin{(A+B)}}{\cos{A} \cos{B}}

3 ( 1 3 + tan ⁡ 1 ° ) ( 1 3 + tan ⁡ 2 ° ) . . . . . . ( 1 3 + tan ⁡ 58 ° ) ( 1 3 + tan ⁡ 59 ° ) \sqrt3 \left(\frac{1}{\sqrt3}+\tan1°\right)\left(\frac{1}{\sqrt3}+\tan2°\right)......\left(\frac{1}{\sqrt3}+\tan58°\right)\left(\frac{1}{\sqrt3}+\tan59°\right)

= 3 ( tan ⁡ 30 ° + tan ⁡ 1 ° ) ( tan ⁡ 30 ° + tan ⁡ 2 ° ) . . . . . . ( tan ⁡ 30 ° + tan ⁡ 58 ° ) ( tan ⁡ 30 ° + tan ⁡ 59 ° ) =\sqrt3 \left(\tan{30°}+\tan1°\right)\left(\tan{30°}+\tan2°\right)......\left(\tan{30°}+\tan58°\right)\left(\tan{30°}+\tan59°\right)

= 3 ( sin ⁡ ( 30 ° + 1 ° ) cos ⁡ 30 ° cos ⁡ 1 ° ) ( sin ⁡ ( 30 ° + 2 ° ) cos ⁡ 30 ° cos ⁡ 2 ° ) . . . . . ( sin ⁡ ( 30 ° + 58 ° ) cos ⁡ 30 ° cos ⁡ 58 ° ) ( sin ⁡ ( 30 ° + 59 ° ) cos ⁡ 30 ° cos ⁡ 59 ° ) =\sqrt{3} \left(\frac{\sin{(30°+1°)}}{\cos{30°} \cos{1°}} \right) \left(\frac{\sin{(30°+2°)}}{\cos{30°} \cos{2°}} \right) ..... \left(\frac{\sin{(30°+58°)}}{\cos{30°} \cos{58°}} \right) \left(\frac{\sin{(30°+59°)}}{\cos{30°} \cos{59°}} \right)

= 3 ( 1 cos ⁡ 30 ° ) 59 ( sin ⁡ 31 ° cos ⁡ 1 ° ) ( sin ⁡ 32 ° cos ⁡ 2 ° ) . . . . ( sin ⁡ 88 ° cos ⁡ 58 ° ) ( sin ⁡ 89 ° cos ⁡ 59 ° ) =\sqrt{3} \left(\frac{1}{\cos{30°}} \right)^{59} \left(\frac{\sin{31°}}{\cos{1°}} \right) \left(\frac{\sin{32°}}{\cos{2°}} \right) .... \left(\frac{\sin{88°}}{\cos{58°}} \right) \left(\frac{\sin{89°}}{\cos{59°}} \right)

We also have: sin ⁡ 90 ° − x ° = cos ⁡ x ° \sin{90°-x°}=\cos{x°} :

. . . = 3 ( 2 3 ) 59 sin ⁡ 31 ° sin ⁡ 32 ° . . . sin ⁡ 88 ° sin ⁡ 89 ° sin ⁡ 89 ° sin ⁡ 88 ° . . . sin ⁡ 32 ° sin ⁡ 31 ° ...= \sqrt{3} \left(\frac{2}{\sqrt{3}} \right)^{59} \frac{\sin{31°} \sin {32°} ... \sin{88°} \sin {89°}}{\sin{89°} \sin{88°} ... \sin{32°} \sin{31°}}

= 3 ( 2 59 ( 3 ) 59 ) = 2 59 ( 3 ) 58 = 2 59 3 29 =\sqrt{3} \left(\frac{2^{59}}{(\sqrt{3})^{59}} \right)= \frac{2^{59}}{(\sqrt{3})^{58}}=\frac{2^{59}}{3^{29}}

a = 2 , b = 59 , c = 3 , d = 29 ⇒ a + b + c + d = 2 + 59 + 3 + 29 = 93 a=2,b=59,c=3,d=29 \Rightarrow a+b+c+d=2+59+3+29=\boxed{93}

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Moderator note:

We can show that ( tan ⁡ 3 0 ∘ + tan ⁡ α ) ( tan ⁡ 3 0 ∘ + tan ⁡ ( 6 0 ∘ − α ) ) = 4 3 (\tan 30 ^ \circ + \tan \alpha) ( \tan 30^\circ + \tan (60^\circ - \alpha) ) = \frac{4}{3} , using the ideas that you've presented.

Very nice proof. Just one thing. I think the denominator of the 6th line is a bit wrong. If its not could you explain that part.

Vignesh S - 5 years, 2 months ago

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Yes sorry, it was a lot wrong. I've fixed it now thank you.

Sam Bealing - 5 years, 2 months ago
Rishabh Jain
Apr 15, 2016

P = 2 ∏ r = 1 29 ( 1 3 + tan ⁡ r ° ) ( 1 3 + tan ⁡ ( 60 − r ) ° ) ⏟ T \mathfrak P=2\prod_{r=1}^{29}\underbrace{\left(\frac{1}{\sqrt 3}+\tan r°\right)\left(\frac{1}{\sqrt 3}+\tan(60- r)°\right)}_{\color{#D61F06}{\mathfrak T}}\

T = 1 / 3 + 1 3 ( tan ⁡ r ° + tan ⁡ ( 60 − r ) ° ) ⏟ A + ( tan ⁡ r ° ) ( tan ⁡ ( 60 − r ° ) ) \color{#D61F06}{\mathfrak T}=1/3+\dfrac{1}{\sqrt 3}\underbrace{(\tan r°+\tan(60-r)°)}_{\large\mathfrak A}+(\tan r°)(\tan(60-r°))

A = tan ⁡ r ° + tan ⁡ ( 60 − r ) ° = ( tan ⁡ r ° + tan ⁡ ( 60 − r ° ) 1 − ( tan ⁡ r ° ) ( tan ⁡ ( 60 − r ° ) ) ( 1 − ( tan ⁡ r ° ) ( tan ⁡ ( 60 − r ° ) ) = 3 ( 1 − ( tan ⁡ r ° ) ( tan ⁡ ( 60 − r ° ) ) \mathfrak A=\tan r°+\tan(60-r)°~~~~~~\\=\left(\color{#3D99F6}{\dfrac{\tan r°+\tan(60-r°)}{1-(\tan r°)(\tan(60-r°)}}\right)(1-(\tan r°)(\tan(60-r°) )\\=\sqrt 3(1-(\tan r°)(\tan(60-r°))~~

∴ T = 4 / 3 \Large\therefore\color{#D61F06}{\mathfrak T}=4/3

P = 2 ∏ r = 1 29 ( 4 3 ) \Large\mathfrak P=2\prod_{r=1}^{29}\left(\dfrac 43\right)

= 2 ( 4 3 ) 29 = 2 59 3 29 \Large =2\left(\dfrac 43\right)^{29}=\dfrac{2^{59}}{3^{29}}

Therefore 2 + 59 + 3 + 29 = 93 2+59+3+29=\boxed{93} .

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I don't know why, but we both have alike solutions many times

Aakash Khandelwal - 5 years, 1 month ago

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Ya... I have this similarity with many of the guys here on brilliant and it's kinda' cool... :-)

Rishabh Jain - 5 years, 1 month ago

Rishabh cool... I just love ur solutions. It motivates me to improve myself..ty

Pawan pal - 5 years, 1 month ago

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Ohhh... Thanks ⌣ ¨ \large\ddot\smile

Rishabh Jain - 5 years, 1 month ago
Ahmad Saad
Apr 16, 2016

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Vignesh S
Apr 15, 2016

  • Combine the terms from the beginning and the end. The general term can be written as ( 1 3 + tan ⁡ ( A ) ) ( 1 3 + tan ⁡ ( 60 ° − A ) ) \left(\dfrac{1}{\sqrt3}+\tan(A)\right)\left(\dfrac{1}{\sqrt3}+\tan(60°-A)\right) = 1 3 + 1 3 ( tan ⁡ ( A ) + tan ⁡ ( 60 ° − A ) + tan ⁡ ( A ) tan ⁡ ( 60 ° − A ) ) =\dfrac{1}{3}+\dfrac{1}{\sqrt3}(\tan(A)+\tan(60°-A)+\tan(A)\tan(60°-A)) = 1 3 + 1 3 ( tan ⁡ ( A + 60 ° − A ) ) ( 1 − tan ⁡ ( A ) tan ⁡ ( 60 ° − A ) ) + tan ⁡ ( A ) tan ⁡ ( 60 ° − A ) =\dfrac{1}{3}+\dfrac{1}{\sqrt3}(\tan(A+60°-A))(1-\tan(A)\tan(60°-A))+\tan(A)\tan(60°-A) = 1 3 + 1 = 4 3 =\dfrac{1}{3}+1=\dfrac{4}{3} But the product term for A = 30 ° A=30° is 2 3 \dfrac{2}{\sqrt3} . We see that for all A ≠ 30 ° A\ne30° the product on multiplication becomes 4 29 3 29 \dfrac{4^{29}}{3^{29}} finally multipying by 2 3 \dfrac{2}{\sqrt3} and 3 \sqrt3 also we get it as 2 59 3 29 \dfrac{2^{59}}{3^{29}} hence the answer is 2 + 3 + 59 + 29 = 93 2+3+59+29=93

    By. g e n e r a l i s e \color{#3D99F6}{generalise} I mean instead of 60 ° 60° take it as some B \color{#3D99F6}{B} and add 1 tan ⁡ ( B ) \color{#D61F06}{\frac{1}{\tan(B)}} to make the product term as ∏ A = 1 ° B − 1 ° ( 1 tan ⁡ ( B ) + tan ⁡ ( B − A ) ) \color{#D61F06}{\Large\prod_{A=1°}^{B-1°}\left(\dfrac{1}{\tan(B)}+\tan(B-A)\right)} and check the middle term once. Try for B = 30 ° \color{#D61F06}{B=30°} the middle term is amazing

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