Find the $frac mn$ if $T=sin 5°+sin10°+sin 15°+cdots+sin175°=tan frac mn$












2












$begingroup$


It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.




$$T=sin(5^circ) +sin(10^circ) + sin(15^circ) + cdots +sin(175^circ) =tan frac mn$$



Find the $frac mn=?$, where $m$ and $n$ are positive integer numbers.




Attepmts:



$$T=2Big(sin(5°)+sin(10°) + cdots + sin(85°)Big) + 1 = 2Big((sin(5°) + cos(5°))+(sin(10°)+ cos(10°))+cdots + (sin(40°)+cos(40°))Big)+1 = 2Big(sqrt 2((sin50°+sin55°)+cdots+sin(80°))Big)+1.$$



and then I can not see an any way...






algebra-precalculus trigonometry summation contest-math







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    See this post to represent the summation as a fraction and continue from there.
    $endgroup$
    – Math Lover
    Jan 2 at 20:18








  • 1




    $begingroup$
    artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/…
    $endgroup$
    – user574848
    Jan 4 at 21:39
















2












$begingroup$


It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.




$$T=sin(5^circ) +sin(10^circ) + sin(15^circ) + cdots +sin(175^circ) =tan frac mn$$



Find the $frac mn=?$, where $m$ and $n$ are positive integer numbers.




Attepmts:



$$T=2Big(sin(5°)+sin(10°) + cdots + sin(85°)Big) + 1 = 2Big((sin(5°) + cos(5°))+(sin(10°)+ cos(10°))+cdots + (sin(40°)+cos(40°))Big)+1 = 2Big(sqrt 2((sin50°+sin55°)+cdots+sin(80°))Big)+1.$$



and then I can not see an any way...






algebra-precalculus trigonometry summation contest-math







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    See this post to represent the summation as a fraction and continue from there.
    $endgroup$
    – Math Lover
    Jan 2 at 20:18








  • 1




    $begingroup$
    artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/…
    $endgroup$
    – user574848
    Jan 4 at 21:39














2












2








2


2



$begingroup$


It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.




$$T=sin(5^circ) +sin(10^circ) + sin(15^circ) + cdots +sin(175^circ) =tan frac mn$$



Find the $frac mn=?$, where $m$ and $n$ are positive integer numbers.




Attepmts:



$$T=2Big(sin(5°)+sin(10°) + cdots + sin(85°)Big) + 1 = 2Big((sin(5°) + cos(5°))+(sin(10°)+ cos(10°))+cdots + (sin(40°)+cos(40°))Big)+1 = 2Big(sqrt 2((sin50°+sin55°)+cdots+sin(80°))Big)+1.$$



and then I can not see an any way...






algebra-precalculus trigonometry summation contest-math







share|cite|improve this question











$endgroup$




It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.




$$T=sin(5^circ) +sin(10^circ) + sin(15^circ) + cdots +sin(175^circ) =tan frac mn$$



Find the $frac mn=?$, where $m$ and $n$ are positive integer numbers.




Attepmts:



$$T=2Big(sin(5°)+sin(10°) + cdots + sin(85°)Big) + 1 = 2Big((sin(5°) + cos(5°))+(sin(10°)+ cos(10°))+cdots + (sin(40°)+cos(40°))Big)+1 = 2Big(sqrt 2((sin50°+sin55°)+cdots+sin(80°))Big)+1.$$



and then I can not see an any way...






algebra-precalculus trigonometry summation contest-math




algebra-precalculus trigonometry summation contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 21:53









El borito

666216




666216










asked Jan 2 at 20:01









ElementaryElementary

361111




361111








  • 4




    $begingroup$
    See this post to represent the summation as a fraction and continue from there.
    $endgroup$
    – Math Lover
    Jan 2 at 20:18








  • 1




    $begingroup$
    artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/…
    $endgroup$
    – user574848
    Jan 4 at 21:39














  • 4




    $begingroup$
    See this post to represent the summation as a fraction and continue from there.
    $endgroup$
    – Math Lover
    Jan 2 at 20:18








  • 1




    $begingroup$
    artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/…
    $endgroup$
    – user574848
    Jan 4 at 21:39








4




4




$begingroup$
See this post to represent the summation as a fraction and continue from there.
$endgroup$
– Math Lover
Jan 2 at 20:18






$begingroup$
See this post to represent the summation as a fraction and continue from there.
$endgroup$
– Math Lover
Jan 2 at 20:18






1




1




$begingroup$
artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/…
$endgroup$
– user574848
Jan 4 at 21:39




$begingroup$
artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/…
$endgroup$
– user574848
Jan 4 at 21:39










2 Answers
2






active

oldest

votes


















6












$begingroup$

Not exactly the most elementary approach, but with complex numbers we could write
$$
T = operatorname{Im}left(1 + e^{frac{pi}{36}i} + e^{2frac{pi}{36}i} + cdots + e^{35frac{pi}{36}i}right) \
=operatorname{Im}left(1 + z + z^2 + cdots + z^{35}right) quad left(z = e^{frac{pi}{36}i}right)\
= operatorname{Im}left(frac{1 - z^{36}}{1 - z}right) = operatorname{Im}left(frac{2}{(1 - cos(5^circ)) - isin(5^circ)}right)\
= operatorname{Im}left(frac{2}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}[(1 - cos(5^circ)) + isin(5^circ)]right)
\ = frac{2 sin(5^circ)}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}
\ = frac{2 sin(5^circ)}{2 - 2cos(5^circ)} = frac{sin(5^circ)}{1 - cos(5^circ)}
$$
As the commenter below points out, we have $displaystylefrac{sin x}{1 - cos x} = tanleft(90^circ - frac{x}{2}right)$. It follows that our answer is
$$
frac{sin(5^circ)}{1 - cos(5^circ)} = tanleft(90^circ - frac{5^circ}{2}right) = tan(87.5^circ) = tanleft(left[frac{175}{2}right]^circright)
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That looks negative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:20










  • $begingroup$
    @LordSharktheUnknown Found the sign error, fixing now
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:21






  • 2




    $begingroup$
    $(sin x)/(1-cos x)=cot (x/2)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:22










  • $begingroup$
    @LordSharktheUnknown Well, that certainly explains it
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:23






  • 1




    $begingroup$
    @Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:32



















3












$begingroup$

By geometric series,
$$T=sin 5°+sin10°+sin 15°+...+sin175° = \
operatorname{Im} (sum_{n=1}^{35}exp(i n 5 pi/180)) =\
operatorname{Im} exp(i 5 pi/180) frac{exp(i 35 cdot 5 pi/180)-1}{exp(i 5 pi/180)-1} =\
operatorname{Im} exp(i 5 pi/180) exp(i 34 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\
operatorname{Im} exp(i 36 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\ frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} \
$$
and
$$
arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( frac{pi}{2} - 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( 35 cdot 5 pi/360)} ) =
frac{35 pi}{72}
$$
or, in degrees, $frac{35 cdot 180}{72} =frac{175}{2} = 87.5 $






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:43










  • $begingroup$
    Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:47










  • $begingroup$
    @Omnomnomnom I put it in the main text
    $endgroup$
    – Andreas
    Jan 2 at 20:53











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Not exactly the most elementary approach, but with complex numbers we could write
$$
T = operatorname{Im}left(1 + e^{frac{pi}{36}i} + e^{2frac{pi}{36}i} + cdots + e^{35frac{pi}{36}i}right) \
=operatorname{Im}left(1 + z + z^2 + cdots + z^{35}right) quad left(z = e^{frac{pi}{36}i}right)\
= operatorname{Im}left(frac{1 - z^{36}}{1 - z}right) = operatorname{Im}left(frac{2}{(1 - cos(5^circ)) - isin(5^circ)}right)\
= operatorname{Im}left(frac{2}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}[(1 - cos(5^circ)) + isin(5^circ)]right)
\ = frac{2 sin(5^circ)}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}
\ = frac{2 sin(5^circ)}{2 - 2cos(5^circ)} = frac{sin(5^circ)}{1 - cos(5^circ)}
$$
As the commenter below points out, we have $displaystylefrac{sin x}{1 - cos x} = tanleft(90^circ - frac{x}{2}right)$. It follows that our answer is
$$
frac{sin(5^circ)}{1 - cos(5^circ)} = tanleft(90^circ - frac{5^circ}{2}right) = tan(87.5^circ) = tanleft(left[frac{175}{2}right]^circright)
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That looks negative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:20










  • $begingroup$
    @LordSharktheUnknown Found the sign error, fixing now
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:21






  • 2




    $begingroup$
    $(sin x)/(1-cos x)=cot (x/2)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:22










  • $begingroup$
    @LordSharktheUnknown Well, that certainly explains it
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:23






  • 1




    $begingroup$
    @Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:32
















6












$begingroup$

Not exactly the most elementary approach, but with complex numbers we could write
$$
T = operatorname{Im}left(1 + e^{frac{pi}{36}i} + e^{2frac{pi}{36}i} + cdots + e^{35frac{pi}{36}i}right) \
=operatorname{Im}left(1 + z + z^2 + cdots + z^{35}right) quad left(z = e^{frac{pi}{36}i}right)\
= operatorname{Im}left(frac{1 - z^{36}}{1 - z}right) = operatorname{Im}left(frac{2}{(1 - cos(5^circ)) - isin(5^circ)}right)\
= operatorname{Im}left(frac{2}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}[(1 - cos(5^circ)) + isin(5^circ)]right)
\ = frac{2 sin(5^circ)}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}
\ = frac{2 sin(5^circ)}{2 - 2cos(5^circ)} = frac{sin(5^circ)}{1 - cos(5^circ)}
$$
As the commenter below points out, we have $displaystylefrac{sin x}{1 - cos x} = tanleft(90^circ - frac{x}{2}right)$. It follows that our answer is
$$
frac{sin(5^circ)}{1 - cos(5^circ)} = tanleft(90^circ - frac{5^circ}{2}right) = tan(87.5^circ) = tanleft(left[frac{175}{2}right]^circright)
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That looks negative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:20










  • $begingroup$
    @LordSharktheUnknown Found the sign error, fixing now
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:21






  • 2




    $begingroup$
    $(sin x)/(1-cos x)=cot (x/2)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:22










  • $begingroup$
    @LordSharktheUnknown Well, that certainly explains it
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:23






  • 1




    $begingroup$
    @Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:32














6












6








6





$begingroup$

Not exactly the most elementary approach, but with complex numbers we could write
$$
T = operatorname{Im}left(1 + e^{frac{pi}{36}i} + e^{2frac{pi}{36}i} + cdots + e^{35frac{pi}{36}i}right) \
=operatorname{Im}left(1 + z + z^2 + cdots + z^{35}right) quad left(z = e^{frac{pi}{36}i}right)\
= operatorname{Im}left(frac{1 - z^{36}}{1 - z}right) = operatorname{Im}left(frac{2}{(1 - cos(5^circ)) - isin(5^circ)}right)\
= operatorname{Im}left(frac{2}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}[(1 - cos(5^circ)) + isin(5^circ)]right)
\ = frac{2 sin(5^circ)}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}
\ = frac{2 sin(5^circ)}{2 - 2cos(5^circ)} = frac{sin(5^circ)}{1 - cos(5^circ)}
$$
As the commenter below points out, we have $displaystylefrac{sin x}{1 - cos x} = tanleft(90^circ - frac{x}{2}right)$. It follows that our answer is
$$
frac{sin(5^circ)}{1 - cos(5^circ)} = tanleft(90^circ - frac{5^circ}{2}right) = tan(87.5^circ) = tanleft(left[frac{175}{2}right]^circright)
$$






share|cite|improve this answer











$endgroup$



Not exactly the most elementary approach, but with complex numbers we could write
$$
T = operatorname{Im}left(1 + e^{frac{pi}{36}i} + e^{2frac{pi}{36}i} + cdots + e^{35frac{pi}{36}i}right) \
=operatorname{Im}left(1 + z + z^2 + cdots + z^{35}right) quad left(z = e^{frac{pi}{36}i}right)\
= operatorname{Im}left(frac{1 - z^{36}}{1 - z}right) = operatorname{Im}left(frac{2}{(1 - cos(5^circ)) - isin(5^circ)}right)\
= operatorname{Im}left(frac{2}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}[(1 - cos(5^circ)) + isin(5^circ)]right)
\ = frac{2 sin(5^circ)}{[1 - cos(5^circ)]^2 + sin^2(5^circ)}
\ = frac{2 sin(5^circ)}{2 - 2cos(5^circ)} = frac{sin(5^circ)}{1 - cos(5^circ)}
$$
As the commenter below points out, we have $displaystylefrac{sin x}{1 - cos x} = tanleft(90^circ - frac{x}{2}right)$. It follows that our answer is
$$
frac{sin(5^circ)}{1 - cos(5^circ)} = tanleft(90^circ - frac{5^circ}{2}right) = tan(87.5^circ) = tanleft(left[frac{175}{2}right]^circright)
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 21:12









El borito

666216




666216










answered Jan 2 at 20:19









OmnomnomnomOmnomnomnom

128k791185




128k791185












  • $begingroup$
    That looks negative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:20










  • $begingroup$
    @LordSharktheUnknown Found the sign error, fixing now
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:21






  • 2




    $begingroup$
    $(sin x)/(1-cos x)=cot (x/2)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:22










  • $begingroup$
    @LordSharktheUnknown Well, that certainly explains it
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:23






  • 1




    $begingroup$
    @Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:32


















  • $begingroup$
    That looks negative.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:20










  • $begingroup$
    @LordSharktheUnknown Found the sign error, fixing now
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:21






  • 2




    $begingroup$
    $(sin x)/(1-cos x)=cot (x/2)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 20:22










  • $begingroup$
    @LordSharktheUnknown Well, that certainly explains it
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:23






  • 1




    $begingroup$
    @Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:32
















$begingroup$
That looks negative.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 20:20




$begingroup$
That looks negative.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 20:20












$begingroup$
@LordSharktheUnknown Found the sign error, fixing now
$endgroup$
– Omnomnomnom
Jan 2 at 20:21




$begingroup$
@LordSharktheUnknown Found the sign error, fixing now
$endgroup$
– Omnomnomnom
Jan 2 at 20:21




2




2




$begingroup$
$(sin x)/(1-cos x)=cot (x/2)$.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 20:22




$begingroup$
$(sin x)/(1-cos x)=cot (x/2)$.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 20:22












$begingroup$
@LordSharktheUnknown Well, that certainly explains it
$endgroup$
– Omnomnomnom
Jan 2 at 20:23




$begingroup$
@LordSharktheUnknown Well, that certainly explains it
$endgroup$
– Omnomnomnom
Jan 2 at 20:23




1




1




$begingroup$
@Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
$endgroup$
– Omnomnomnom
Jan 2 at 20:32




$begingroup$
@Beginner you should see this post linked above. It looks to me like you were at least trying the right kind of trick
$endgroup$
– Omnomnomnom
Jan 2 at 20:32











3












$begingroup$

By geometric series,
$$T=sin 5°+sin10°+sin 15°+...+sin175° = \
operatorname{Im} (sum_{n=1}^{35}exp(i n 5 pi/180)) =\
operatorname{Im} exp(i 5 pi/180) frac{exp(i 35 cdot 5 pi/180)-1}{exp(i 5 pi/180)-1} =\
operatorname{Im} exp(i 5 pi/180) exp(i 34 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\
operatorname{Im} exp(i 36 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\ frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} \
$$
and
$$
arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( frac{pi}{2} - 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( 35 cdot 5 pi/360)} ) =
frac{35 pi}{72}
$$
or, in degrees, $frac{35 cdot 180}{72} =frac{175}{2} = 87.5 $






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:43










  • $begingroup$
    Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:47










  • $begingroup$
    @Omnomnomnom I put it in the main text
    $endgroup$
    – Andreas
    Jan 2 at 20:53
















3












$begingroup$

By geometric series,
$$T=sin 5°+sin10°+sin 15°+...+sin175° = \
operatorname{Im} (sum_{n=1}^{35}exp(i n 5 pi/180)) =\
operatorname{Im} exp(i 5 pi/180) frac{exp(i 35 cdot 5 pi/180)-1}{exp(i 5 pi/180)-1} =\
operatorname{Im} exp(i 5 pi/180) exp(i 34 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\
operatorname{Im} exp(i 36 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\ frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} \
$$
and
$$
arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( frac{pi}{2} - 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( 35 cdot 5 pi/360)} ) =
frac{35 pi}{72}
$$
or, in degrees, $frac{35 cdot 180}{72} =frac{175}{2} = 87.5 $






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:43










  • $begingroup$
    Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:47










  • $begingroup$
    @Omnomnomnom I put it in the main text
    $endgroup$
    – Andreas
    Jan 2 at 20:53














3












3








3





$begingroup$

By geometric series,
$$T=sin 5°+sin10°+sin 15°+...+sin175° = \
operatorname{Im} (sum_{n=1}^{35}exp(i n 5 pi/180)) =\
operatorname{Im} exp(i 5 pi/180) frac{exp(i 35 cdot 5 pi/180)-1}{exp(i 5 pi/180)-1} =\
operatorname{Im} exp(i 5 pi/180) exp(i 34 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\
operatorname{Im} exp(i 36 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\ frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} \
$$
and
$$
arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( frac{pi}{2} - 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( 35 cdot 5 pi/360)} ) =
frac{35 pi}{72}
$$
or, in degrees, $frac{35 cdot 180}{72} =frac{175}{2} = 87.5 $






share|cite|improve this answer











$endgroup$



By geometric series,
$$T=sin 5°+sin10°+sin 15°+...+sin175° = \
operatorname{Im} (sum_{n=1}^{35}exp(i n 5 pi/180)) =\
operatorname{Im} exp(i 5 pi/180) frac{exp(i 35 cdot 5 pi/180)-1}{exp(i 5 pi/180)-1} =\
operatorname{Im} exp(i 5 pi/180) exp(i 34 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\
operatorname{Im} exp(i 36 cdot 5 pi/360) frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} =
\ frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} \
$$
and
$$
arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( frac{pi}{2} - 5 pi/360)} ) = \
arctan(frac{sin( 35 cdot 5 pi/360)}{cos( 35 cdot 5 pi/360)} ) =
frac{35 pi}{72}
$$
or, in degrees, $frac{35 cdot 180}{72} =frac{175}{2} = 87.5 $







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 20:59

























answered Jan 2 at 20:38









AndreasAndreas

8,1481137




8,1481137












  • $begingroup$
    Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:43










  • $begingroup$
    Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:47










  • $begingroup$
    @Omnomnomnom I put it in the main text
    $endgroup$
    – Andreas
    Jan 2 at 20:53


















  • $begingroup$
    Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:43










  • $begingroup$
    Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
    $endgroup$
    – Omnomnomnom
    Jan 2 at 20:47










  • $begingroup$
    @Omnomnomnom I put it in the main text
    $endgroup$
    – Andreas
    Jan 2 at 20:53
















$begingroup$
Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
$endgroup$
– Omnomnomnom
Jan 2 at 20:43




$begingroup$
Is there a quick or clever way to compute $$ arctan(frac{sin( 35 cdot 5 pi/360)}{sin( 5 pi/360)} ) = frac{35 pi}{72} ? $$
$endgroup$
– Omnomnomnom
Jan 2 at 20:43












$begingroup$
Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
$endgroup$
– Omnomnomnom
Jan 2 at 20:47




$begingroup$
Never mind, I see now that the expression in the parentheses can immediately be rewritten as $sin(87.5^circ)/cos(87.5^circ)$.
$endgroup$
– Omnomnomnom
Jan 2 at 20:47












$begingroup$
@Omnomnomnom I put it in the main text
$endgroup$
– Andreas
Jan 2 at 20:53




$begingroup$
@Omnomnomnom I put it in the main text
$endgroup$
– Andreas
Jan 2 at 20:53


















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