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Geometry Level 2

∑ k = 1 50 [ ( 1 + tan ⁡ ( k ∘ ) + sec ⁡ ( k ∘ ) ) ( 1 + cot ⁡ ( k ∘ ) − csc ⁡ ( k ∘ ) ) ] = ? \displaystyle \sum_{k=1}^{50} \Bigg [ \bigg(1 + \tan(k^\circ)+\sec(k^\circ)\bigg)\ \bigg(1+\cot(k^\circ)-\csc(k^\circ) \bigg)\Bigg]=\ ? \


The answer is 100.

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K. J. W. by K. J. W. , 20, Singapore

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1 solution

K. J. W.
May 24, 2015

Note that ( 1 + tan ⁡ k ° + sec ⁡ k ° ) ( 1 + cot ⁡ k ° − csc ⁡ k ° ) = 1 cos ⁡ k ° ( cos ⁡ k ° + sin ⁡ k ° + 1 ) 1 sin ⁡ k ° ( sin ⁡ k ° + cos ⁡ k ° − 1 ) = 1 cos ⁡ k ° sin ⁡ k ° [ ( sin ⁡ k ° + cos ⁡ k ° ) 2 − 1 2 ] = 1 cos ⁡ k ° sin ⁡ k ° ( sin ⁡ 2 k ° + 2 cos ⁡ k ° sin ⁡ k ° + cos ⁡ 2 k ° − 1 ) = 1 cos ⁡ k ° sin ⁡ k ° ( 2 cos ⁡ k ° sin ⁡ k ° ) = 2 ∴ Sum = ∑ k = 1 50 2 = 100 . { (1+\tan k°+\sec k°)(1+\cot k°-\csc k°) }\\ =\dfrac { 1 }{ \cos k° } (\cos k°+\sin k°+1)\dfrac { 1 }{ \sin k° } (\sin k°+\cos k°-1)\\ =\dfrac { 1 }{ \cos k°\sin k° } \left[ { \left( \sin k°+\cos k° \right) }^{ 2 }-{ 1 }^{ 2 } \right] \\ =\dfrac { 1 }{ \cos k°\sin k° } (\sin ^{ 2 }{ k°+2\cos k°\sin k°+\cos ^{ 2 }{ k°-1) } } \\ =\dfrac { 1 }{ \cos k°\sin k° } (2\cos k°\sin k°)\\ =2\\ \therefore \quad \text{Sum}=\displaystyle \sum _{ k=1 }^{ 50 }{ 2 } =\boxed { 100 } .

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Moderator note:

Nice. Bonus question: What would the answer be if I replace the number 50 50 by 100 100 ?

Actually there is no double angle formula involved.

Formulae involved: s i n ( k ° ) cos ⁡ ( k ° ) = t a n ( k ° ) , sin ⁡ 2 ( k ° ) + cos ⁡ 2 ( k ° ) = 1 \frac { sin(k°) }{ \cos { (k°) } } =tan(k°),\quad \sin ^{ 2 }{ (k°)+\cos ^{ 2 }{ (k°)=1 } }

BTW, if 50 is replaced with 100, the answer would be undefined as t a n ( 90 ° ) tan(90°) is undefined and 90 is a term in the summation.

K. J. W. - 6 years ago

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