Logarithm + Trigonometry = Summation

Geometry Level 4

log ⁡ 2 ( 1 + cot ⁡ 4 6 ∘ ) + log ⁡ 2 ( 1 + cot ⁡ 4 7 ∘ ) + log ⁡ 2 ( 1 + cot ⁡ 4 8 ∘ ) + … + log ⁡ 2 ( 1 + cot ⁡ 8 9 ∘ ) = ? \begin{aligned} \large &&\log_{2}{(1+\cot{46^{\circ}})} \\ \large&+&\log_{2}{(1+\cot{47^{\circ}})}\\ \large &+&\log_{2}{(1+\cot{48^{\circ}})} \\ \large &+&\ldots \\ \large &+&\log_{2}{(1+\cot{89^{\circ}})}\\ \large &=&\ ? \end{aligned}


The answer is 22.

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Ikkyu San by Ikkyu San , 23, Thailand

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3 solutions

1 + cot ⁡ ( A ) = 2 sin ⁡ ( 4 5 ∘ + A ) sin ⁡ ( A ) 1+\cot(A)=\sqrt{2} \frac{\sin(45^\circ +A)}{\sin(A)}

Using the property of logarithms: log ⁡ a ( x ) + log ⁡ a ( y ) = log ⁡ a ( x y ) \log_a(x)+\log_a(y)=\log_a(xy) :

Answer= log ⁡ 2 ( 2 22 × sin ⁡ ( 4 5 ∘ + 4 6 ∘ ) sin ⁡ ( 4 5 ∘ + 4 7 ∘ ) . . . sin ⁡ ( 4 5 ∘ + 8 9 ∘ ) sin ⁡ ( 4 6 ∘ ) sin ⁡ ( 4 7 ∘ ) . . . sin ⁡ ( 8 9 ∘ ) ) \log_2\left(2^{22} \times \frac{\sin(45^\circ +46^\circ )\sin(45^\circ +47^\circ )...\sin(45^\circ +89^\circ )}{\sin(46^\circ )\sin(47^\circ )...\sin(89^\circ)}\right)

Using the identity sin ⁡ ( 9 0 ∘ + A ) = cos ⁡ ( A ) \sin(90^\circ + A)=\cos(A) in numerator, we have:

= log ⁡ 2 ( 2 22 × cos ⁡ ( 1 ∘ ) cos ⁡ ( 2 ∘ ) . . . cos ⁡ ( 4 4 ∘ ) sin ⁡ ( 4 6 ∘ ) sin ⁡ ( 4 7 ∘ ) . . . sin ⁡ ( 8 9 ∘ ) ) =\log_2\left(2^{22} \times \frac{\cos(1^\circ)\cos(2^\circ)...\cos(44^\circ )}{\sin(46^\circ )\sin(47^\circ )...\sin(89^\circ)}\right)

Using the identity sin ⁡ ( 9 0 ∘ − A ) = cos ⁡ ( A ) \sin(90^\circ-A)=\cos(A) in denominator, we have:

= log ⁡ 2 ( 2 22 × cos ⁡ ( 1 ∘ ) cos ⁡ ( 2 ∘ ) . . . cos ⁡ ( 4 4 ∘ ) cos ⁡ ( 4 4 ∘ ) cos ⁡ ( 4 3 ∘ ) . . . cos ⁡ ( 1 ∘ ) ) =\log_2\left(2^{22} \times \frac{\cos(1^\circ)\cos(2^\circ)...\cos(44^\circ )}{\cos(44^\circ )\cos(43^\circ )...\cos(1^\circ)}\right)

= log ⁡ 2 ( 2 22 ) = 22 =\log_2(2^{22})=22

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@Raghav Vaidyanathan , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 6 years ago

Using the log property log ⁡ a ( x ) + log ⁡ a ( y ) = log ⁡ a ( x y ) \log_{a}(x) + \log_{a}(y) = \log_{a}(xy) in any (valid) base a a , and the fact that cot ⁡ ( k ∘ ) = tan ⁡ ( 9 0 ∘ − k ∘ ) , \cot(k^{\circ}) = \tan(90^{\circ} - k^{\circ}), this expression can be written as

log ⁡ 2 ∏ k = 1 44 ( 1 + tan ⁡ ( k ∘ ) ) , \log_{2}\displaystyle\prod_{k=1}^{44} (1 + \tan(k^{\circ})), (A).

Now 1 = tan ⁡ ( 4 5 ∘ ) = tan ⁡ ( k ∘ ) + tan ⁡ ( 4 5 ∘ − k ∘ ) 1 − tan ⁡ ( k ∘ ) tan ⁡ ( 4 5 ∘ − k ∘ ) 1 = \tan(45^{\circ}) = \dfrac{\tan(k^{\circ}) + \tan(45^{\circ} - k^{\circ})}{1 - \tan(k^{\circ})\tan(45^{\circ} - k^{\circ})}

⟹ tan ⁡ ( k ∘ ) + tan ⁡ ( 4 5 ∘ − k ∘ ) = 1 − tan ⁡ ( k ∘ ) tan ⁡ ( 4 5 ∘ − k ∘ ) , \Longrightarrow \tan(k^{\circ}) + \tan(45^{\circ} - k^{\circ}) = 1 - \tan(k^{\circ})\tan(45^{\circ} - k^{\circ}), (B).

Next, we have ( 1 + tan ⁡ ( k ∘ ) ( 1 + tan ⁡ ( 4 5 ∘ ) ) = (1 + \tan(k^{\circ})(1 + \tan(45^{\circ})) =

1 + tan ⁡ ( k ∘ ) + tan ⁡ ( 4 5 ∘ ) + tan ⁡ ( k ∘ ) tan ⁡ ( 4 5 ∘ − k ∘ ) = 1 + 1 = 2 , 1 + \tan(k^{\circ}) + \tan(45^{\circ}) + \tan(k^{\circ})\tan(45^{\circ} - k^{\circ}) = 1 + 1 = 2,

where we have substituted in equation (B). We have 22 22 such multiplicative pairs in our product (A), giving us the answer log ⁡ 2 ( 2 22 ) = 22 . \log_{2}(2^{22}) = \boxed{22}.

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Alternatively:

1 + c o t ( A ) = 2 s i n ( 4 5 o + A ) s i n ( A ) 1+cot(A)=\sqrt{2} \frac{sin(45^o +A)}{sin(A)} ]

Using the property of logarithms: l o g a ( x ) + l o g a ( y ) = l o g a ( x y ) log_a(x)+log_a(y)=log_a(xy) :

Answer= l o g 2 ( 2 22 × s i n ( 4 5 o + 4 6 o ) s i n ( 4 5 o + 4 7 o ) . . . s i n ( 4 5 o + 8 9 o ) s i n ( 4 6 o ) s i n ( 4 7 o ) . . . s i n ( 8 9 o ) ) log_2(2^{22} \times \frac{sin(45^o +46^o )sin(45^o +47^o )...sin(45^o +89^o )}{sin(46^o )sin(47^o )...sin(89^o)})

Using the identity s i n ( 9 0 o + A ) = c o s ( A ) sin(90^o + A)=cos(A) in numerator, we have:

= l o g 2 ( 2 22 × c o s ( 1 o ) c o s ( 2 o ) . . . c o s ( 4 4 o ) s i n ( 4 6 o ) s i n ( 4 7 o ) . . . s i n ( 8 9 o ) ) =log_2(2^{22} \times \frac{cos(1^o)cos(2^o)...cos(44^o )}{sin(46^o )sin(47^o )...sin(89^o)})

Using the identity s i n ( 9 0 o − A ) = c o s ( A ) sin(90^o-A)=cos(A) in denominator, we have:

= l o g 2 ( 2 22 × c o s ( 1 o ) c o s ( 2 o ) . . . c o s ( 4 4 o ) c o s ( 4 4 o ) c o s ( 4 3 o ) . . . c o s ( 1 o ) ) =log_2(2^{22} \times \frac{cos(1^o)cos(2^o)...cos(44^o )}{cos(44^o )cos(43^o )...cos(1^o)})

= l o g 2 ( 2 22 ) = 22 =log_2(2^{22})=22

Raghav Vaidyanathan - 6 years ago

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i have another way:

if A+B=135 then (1+cotA)(1+cotB)=2

then using logx+logy=logxy

then xy=(1+cot46)(1+cot47)........(1+cot88)(1+cot89)=2 to the power 22

then answer is i.e; logxy to the base 2=22

Bharat Naik - 6 years ago

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That is a great observation! +1

Raghav Vaidyanathan - 6 years ago

Nice approach. I hadn't seen that initial identity before. In that last step it might have been better to convert sin ⁡ ( 46 ) sin ⁡ ( 47 ) . . . sin ⁡ ( 89 ) \sin(46)\sin(47)...\sin(89) to cos ⁡ ( 1 ) cos ⁡ ( 2 ) . . . . cos ⁡ ( 44 ) \cos(1)\cos(2)....\cos(44) to make the fact that the trig fraction is 1 1 more apparent. :)

Brian Charlesworth - 6 years ago

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I made a mistake, I will correct it. The last step is incorrect.

Raghav Vaidyanathan - 6 years ago

tan45.tank=tank.tan(45-k) ?

Soner Karaca - 6 years ago
Cs ಠ_ಠ Lee
May 24, 2015

Insert Raghave Vaidyanathan's pro solution here

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