If $sin 5°+sin 10°+sin15°+cdots+sin 40°=a$, then $sin 5°+sin 10°+sin15°+cdots+sin 175°=?$












5












$begingroup$


I'm stuck in this question




If $sin 5°+sin 10°+sin15°+cdots+sin 40°=a$



$sin 5°+sin 10°+sin15°+cdots+sin 175°=?$




I know that, (I asked before) $sin 5°+sin 10°+sin15°+cdots+sin 175°=tanfrac{175}{2}$



But, I didn't catch a hint here.






algebra-precalculus trigonometry summation contest-math







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
    $endgroup$
    – Thomas Shelby
    Jan 4 at 12:20












  • $begingroup$
    You can link to your question more easier than anybody.
    $endgroup$
    – kelalaka
    Jan 4 at 15:02










  • $begingroup$
    math.stackexchange.com/questions/17966/…
    $endgroup$
    – lab bhattacharjee
    Jan 4 at 19:47
















5












$begingroup$


I'm stuck in this question




If $sin 5°+sin 10°+sin15°+cdots+sin 40°=a$



$sin 5°+sin 10°+sin15°+cdots+sin 175°=?$




I know that, (I asked before) $sin 5°+sin 10°+sin15°+cdots+sin 175°=tanfrac{175}{2}$



But, I didn't catch a hint here.






algebra-precalculus trigonometry summation contest-math







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
    $endgroup$
    – Thomas Shelby
    Jan 4 at 12:20












  • $begingroup$
    You can link to your question more easier than anybody.
    $endgroup$
    – kelalaka
    Jan 4 at 15:02










  • $begingroup$
    math.stackexchange.com/questions/17966/…
    $endgroup$
    – lab bhattacharjee
    Jan 4 at 19:47














5












5








5


0



$begingroup$


I'm stuck in this question




If $sin 5°+sin 10°+sin15°+cdots+sin 40°=a$



$sin 5°+sin 10°+sin15°+cdots+sin 175°=?$




I know that, (I asked before) $sin 5°+sin 10°+sin15°+cdots+sin 175°=tanfrac{175}{2}$



But, I didn't catch a hint here.






algebra-precalculus trigonometry summation contest-math







share|cite|improve this question











$endgroup$




I'm stuck in this question




If $sin 5°+sin 10°+sin15°+cdots+sin 40°=a$



$sin 5°+sin 10°+sin15°+cdots+sin 175°=?$




I know that, (I asked before) $sin 5°+sin 10°+sin15°+cdots+sin 175°=tanfrac{175}{2}$



But, I didn't catch a hint here.






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 4 at 11:29









Blue

48.1k870153




48.1k870153










asked Jan 4 at 11:22









ElementaryElementary

364110




364110








  • 1




    $begingroup$
    $sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
    $endgroup$
    – Thomas Shelby
    Jan 4 at 12:20












  • $begingroup$
    You can link to your question more easier than anybody.
    $endgroup$
    – kelalaka
    Jan 4 at 15:02










  • $begingroup$
    math.stackexchange.com/questions/17966/…
    $endgroup$
    – lab bhattacharjee
    Jan 4 at 19:47














  • 1




    $begingroup$
    $sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
    $endgroup$
    – Thomas Shelby
    Jan 4 at 12:20












  • $begingroup$
    You can link to your question more easier than anybody.
    $endgroup$
    – kelalaka
    Jan 4 at 15:02










  • $begingroup$
    math.stackexchange.com/questions/17966/…
    $endgroup$
    – lab bhattacharjee
    Jan 4 at 19:47








1




1




$begingroup$
$sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
$endgroup$
– Thomas Shelby
Jan 4 at 12:20






$begingroup$
$sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
$endgroup$
– Thomas Shelby
Jan 4 at 12:20














$begingroup$
You can link to your question more easier than anybody.
$endgroup$
– kelalaka
Jan 4 at 15:02




$begingroup$
You can link to your question more easier than anybody.
$endgroup$
– kelalaka
Jan 4 at 15:02












$begingroup$
math.stackexchange.com/questions/17966/…
$endgroup$
– lab bhattacharjee
Jan 4 at 19:47




$begingroup$
math.stackexchange.com/questions/17966/…
$endgroup$
– lab bhattacharjee
Jan 4 at 19:47










2 Answers
2






active

oldest

votes


















3












$begingroup$

Let
$$
a = sin 5°+sin 10°+sin15°+cdots+sin 40°= \
operatorname{Im} (sum_{n=0}^{8}exp(i n 5 pi/180)) $$
and
$$
b = sin 5°+sin 10°+sin15°+cdots+sin 175°= \
operatorname{Im} left((1 + exp(i 9cdot 5 pi/180)+ exp(i 2cdot 9cdot 5 pi/180)+ exp(i 3cdot9cdot 5 pi/180))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + exp(i pi/4)+ exp(i 2 pi/4)+ exp(i 3 pi/4))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + i(1 +sqrt 2))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
a + (1 +sqrt 2)operatorname{Re} left( sum_{n=0}^{8}exp(i n 5 pi/180)right) = a + (1 +sqrt 2)sum_{n=0}^{8}cos( n 5 pi/180)\
= a + (1 +sqrt 2)sum_{n=0}^{8}sin((90 - 5 n) pi/180)
$$
Denote $c = sum_{n=0}^{8}sin((90 - 5 n) pi/180)$. Then we have
$$
b = 2 a + 2c +frac{2}{sqrt 2} -1
$$
The last two terms are $sin 45° = frac{1}{sqrt 2}$, $sin 135° = frac{1}{sqrt 2}$. And $sin 90° = 1$ has to be subtracted because it was counted twice in $2c$.



Solving $$
b = 2 a + 2c +frac{2}{sqrt 2} -1 \
b = a + (1 +sqrt 2) c
$$



gives $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) Many thanks. I hope we can complete.
    $endgroup$
    – Elementary
    Jan 4 at 12:19










  • $begingroup$
    $frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
    $endgroup$
    – Aleksas Domarkas
    Jan 4 at 12:43






  • 1




    $begingroup$
    @Beginner I completed that approach.
    $endgroup$
    – Andreas
    Jan 4 at 12:59



















1












$begingroup$

Using How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?,



$2acdotsin2.5^circ=cos2.5^circ-cos(45^circ-2.5^circ)$



$iffleft(2a+dfrac1{sqrt2}right)sin2.5^circ=left(1-dfrac1{sqrt2}right)cos2.5^circ (1)$



If $b=sin 5^circ+sin 10^circ+sin15^circ+cdots+sin 175^circ,$



$2bcdotsin2.5^circ=cos2.5^circ-cos(180^circ-2.5^circ)=2cos2.5^circ (2)$



Divide $(2)$ by $(1)$ to find $$dfrac{2b}{left(2a+dfrac1{sqrt2}right)}=dfrac2{left(1-dfrac1{sqrt2}right)}$$ as $sin2.5^circcdotcos2.5^circne0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer.
    $endgroup$
    – Elementary
    Jan 5 at 14:03











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Let
$$
a = sin 5°+sin 10°+sin15°+cdots+sin 40°= \
operatorname{Im} (sum_{n=0}^{8}exp(i n 5 pi/180)) $$
and
$$
b = sin 5°+sin 10°+sin15°+cdots+sin 175°= \
operatorname{Im} left((1 + exp(i 9cdot 5 pi/180)+ exp(i 2cdot 9cdot 5 pi/180)+ exp(i 3cdot9cdot 5 pi/180))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + exp(i pi/4)+ exp(i 2 pi/4)+ exp(i 3 pi/4))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + i(1 +sqrt 2))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
a + (1 +sqrt 2)operatorname{Re} left( sum_{n=0}^{8}exp(i n 5 pi/180)right) = a + (1 +sqrt 2)sum_{n=0}^{8}cos( n 5 pi/180)\
= a + (1 +sqrt 2)sum_{n=0}^{8}sin((90 - 5 n) pi/180)
$$
Denote $c = sum_{n=0}^{8}sin((90 - 5 n) pi/180)$. Then we have
$$
b = 2 a + 2c +frac{2}{sqrt 2} -1
$$
The last two terms are $sin 45° = frac{1}{sqrt 2}$, $sin 135° = frac{1}{sqrt 2}$. And $sin 90° = 1$ has to be subtracted because it was counted twice in $2c$.



Solving $$
b = 2 a + 2c +frac{2}{sqrt 2} -1 \
b = a + (1 +sqrt 2) c
$$



gives $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) Many thanks. I hope we can complete.
    $endgroup$
    – Elementary
    Jan 4 at 12:19










  • $begingroup$
    $frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
    $endgroup$
    – Aleksas Domarkas
    Jan 4 at 12:43






  • 1




    $begingroup$
    @Beginner I completed that approach.
    $endgroup$
    – Andreas
    Jan 4 at 12:59
















3












$begingroup$

Let
$$
a = sin 5°+sin 10°+sin15°+cdots+sin 40°= \
operatorname{Im} (sum_{n=0}^{8}exp(i n 5 pi/180)) $$
and
$$
b = sin 5°+sin 10°+sin15°+cdots+sin 175°= \
operatorname{Im} left((1 + exp(i 9cdot 5 pi/180)+ exp(i 2cdot 9cdot 5 pi/180)+ exp(i 3cdot9cdot 5 pi/180))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + exp(i pi/4)+ exp(i 2 pi/4)+ exp(i 3 pi/4))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + i(1 +sqrt 2))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
a + (1 +sqrt 2)operatorname{Re} left( sum_{n=0}^{8}exp(i n 5 pi/180)right) = a + (1 +sqrt 2)sum_{n=0}^{8}cos( n 5 pi/180)\
= a + (1 +sqrt 2)sum_{n=0}^{8}sin((90 - 5 n) pi/180)
$$
Denote $c = sum_{n=0}^{8}sin((90 - 5 n) pi/180)$. Then we have
$$
b = 2 a + 2c +frac{2}{sqrt 2} -1
$$
The last two terms are $sin 45° = frac{1}{sqrt 2}$, $sin 135° = frac{1}{sqrt 2}$. And $sin 90° = 1$ has to be subtracted because it was counted twice in $2c$.



Solving $$
b = 2 a + 2c +frac{2}{sqrt 2} -1 \
b = a + (1 +sqrt 2) c
$$



gives $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    (+1) Many thanks. I hope we can complete.
    $endgroup$
    – Elementary
    Jan 4 at 12:19










  • $begingroup$
    $frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
    $endgroup$
    – Aleksas Domarkas
    Jan 4 at 12:43






  • 1




    $begingroup$
    @Beginner I completed that approach.
    $endgroup$
    – Andreas
    Jan 4 at 12:59














3












3








3





$begingroup$

Let
$$
a = sin 5°+sin 10°+sin15°+cdots+sin 40°= \
operatorname{Im} (sum_{n=0}^{8}exp(i n 5 pi/180)) $$
and
$$
b = sin 5°+sin 10°+sin15°+cdots+sin 175°= \
operatorname{Im} left((1 + exp(i 9cdot 5 pi/180)+ exp(i 2cdot 9cdot 5 pi/180)+ exp(i 3cdot9cdot 5 pi/180))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + exp(i pi/4)+ exp(i 2 pi/4)+ exp(i 3 pi/4))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + i(1 +sqrt 2))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
a + (1 +sqrt 2)operatorname{Re} left( sum_{n=0}^{8}exp(i n 5 pi/180)right) = a + (1 +sqrt 2)sum_{n=0}^{8}cos( n 5 pi/180)\
= a + (1 +sqrt 2)sum_{n=0}^{8}sin((90 - 5 n) pi/180)
$$
Denote $c = sum_{n=0}^{8}sin((90 - 5 n) pi/180)$. Then we have
$$
b = 2 a + 2c +frac{2}{sqrt 2} -1
$$
The last two terms are $sin 45° = frac{1}{sqrt 2}$, $sin 135° = frac{1}{sqrt 2}$. And $sin 90° = 1$ has to be subtracted because it was counted twice in $2c$.



Solving $$
b = 2 a + 2c +frac{2}{sqrt 2} -1 \
b = a + (1 +sqrt 2) c
$$



gives $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$






share|cite|improve this answer











$endgroup$



Let
$$
a = sin 5°+sin 10°+sin15°+cdots+sin 40°= \
operatorname{Im} (sum_{n=0}^{8}exp(i n 5 pi/180)) $$
and
$$
b = sin 5°+sin 10°+sin15°+cdots+sin 175°= \
operatorname{Im} left((1 + exp(i 9cdot 5 pi/180)+ exp(i 2cdot 9cdot 5 pi/180)+ exp(i 3cdot9cdot 5 pi/180))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + exp(i pi/4)+ exp(i 2 pi/4)+ exp(i 3 pi/4))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
operatorname{Im} left((1 + i(1 +sqrt 2))cdot sum_{n=0}^{8}exp(i n 5 pi/180)right) =\
a + (1 +sqrt 2)operatorname{Re} left( sum_{n=0}^{8}exp(i n 5 pi/180)right) = a + (1 +sqrt 2)sum_{n=0}^{8}cos( n 5 pi/180)\
= a + (1 +sqrt 2)sum_{n=0}^{8}sin((90 - 5 n) pi/180)
$$
Denote $c = sum_{n=0}^{8}sin((90 - 5 n) pi/180)$. Then we have
$$
b = 2 a + 2c +frac{2}{sqrt 2} -1
$$
The last two terms are $sin 45° = frac{1}{sqrt 2}$, $sin 135° = frac{1}{sqrt 2}$. And $sin 90° = 1$ has to be subtracted because it was counted twice in $2c$.



Solving $$
b = 2 a + 2c +frac{2}{sqrt 2} -1 \
b = a + (1 +sqrt 2) c
$$



gives $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 13:32

























answered Jan 4 at 12:05









AndreasAndreas

8,1531037




8,1531037












  • $begingroup$
    (+1) Many thanks. I hope we can complete.
    $endgroup$
    – Elementary
    Jan 4 at 12:19










  • $begingroup$
    $frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
    $endgroup$
    – Aleksas Domarkas
    Jan 4 at 12:43






  • 1




    $begingroup$
    @Beginner I completed that approach.
    $endgroup$
    – Andreas
    Jan 4 at 12:59


















  • $begingroup$
    (+1) Many thanks. I hope we can complete.
    $endgroup$
    – Elementary
    Jan 4 at 12:19










  • $begingroup$
    $frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
    $endgroup$
    – Aleksas Domarkas
    Jan 4 at 12:43






  • 1




    $begingroup$
    @Beginner I completed that approach.
    $endgroup$
    – Andreas
    Jan 4 at 12:59
















$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Elementary
Jan 4 at 12:19




$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Elementary
Jan 4 at 12:19












$begingroup$
$frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
$endgroup$
– Aleksas Domarkas
Jan 4 at 12:43




$begingroup$
$frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
$endgroup$
– Aleksas Domarkas
Jan 4 at 12:43




1




1




$begingroup$
@Beginner I completed that approach.
$endgroup$
– Andreas
Jan 4 at 12:59




$begingroup$
@Beginner I completed that approach.
$endgroup$
– Andreas
Jan 4 at 12:59











1












$begingroup$

Using How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?,



$2acdotsin2.5^circ=cos2.5^circ-cos(45^circ-2.5^circ)$



$iffleft(2a+dfrac1{sqrt2}right)sin2.5^circ=left(1-dfrac1{sqrt2}right)cos2.5^circ (1)$



If $b=sin 5^circ+sin 10^circ+sin15^circ+cdots+sin 175^circ,$



$2bcdotsin2.5^circ=cos2.5^circ-cos(180^circ-2.5^circ)=2cos2.5^circ (2)$



Divide $(2)$ by $(1)$ to find $$dfrac{2b}{left(2a+dfrac1{sqrt2}right)}=dfrac2{left(1-dfrac1{sqrt2}right)}$$ as $sin2.5^circcdotcos2.5^circne0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer.
    $endgroup$
    – Elementary
    Jan 5 at 14:03
















1












$begingroup$

Using How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?,



$2acdotsin2.5^circ=cos2.5^circ-cos(45^circ-2.5^circ)$



$iffleft(2a+dfrac1{sqrt2}right)sin2.5^circ=left(1-dfrac1{sqrt2}right)cos2.5^circ (1)$



If $b=sin 5^circ+sin 10^circ+sin15^circ+cdots+sin 175^circ,$



$2bcdotsin2.5^circ=cos2.5^circ-cos(180^circ-2.5^circ)=2cos2.5^circ (2)$



Divide $(2)$ by $(1)$ to find $$dfrac{2b}{left(2a+dfrac1{sqrt2}right)}=dfrac2{left(1-dfrac1{sqrt2}right)}$$ as $sin2.5^circcdotcos2.5^circne0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer.
    $endgroup$
    – Elementary
    Jan 5 at 14:03














1












1








1





$begingroup$

Using How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?,



$2acdotsin2.5^circ=cos2.5^circ-cos(45^circ-2.5^circ)$



$iffleft(2a+dfrac1{sqrt2}right)sin2.5^circ=left(1-dfrac1{sqrt2}right)cos2.5^circ (1)$



If $b=sin 5^circ+sin 10^circ+sin15^circ+cdots+sin 175^circ,$



$2bcdotsin2.5^circ=cos2.5^circ-cos(180^circ-2.5^circ)=2cos2.5^circ (2)$



Divide $(2)$ by $(1)$ to find $$dfrac{2b}{left(2a+dfrac1{sqrt2}right)}=dfrac2{left(1-dfrac1{sqrt2}right)}$$ as $sin2.5^circcdotcos2.5^circne0$






share|cite|improve this answer









$endgroup$



Using How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?,



$2acdotsin2.5^circ=cos2.5^circ-cos(45^circ-2.5^circ)$



$iffleft(2a+dfrac1{sqrt2}right)sin2.5^circ=left(1-dfrac1{sqrt2}right)cos2.5^circ (1)$



If $b=sin 5^circ+sin 10^circ+sin15^circ+cdots+sin 175^circ,$



$2bcdotsin2.5^circ=cos2.5^circ-cos(180^circ-2.5^circ)=2cos2.5^circ (2)$



Divide $(2)$ by $(1)$ to find $$dfrac{2b}{left(2a+dfrac1{sqrt2}right)}=dfrac2{left(1-dfrac1{sqrt2}right)}$$ as $sin2.5^circcdotcos2.5^circne0$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 13:25









lab bhattacharjeelab bhattacharjee

225k15157275




225k15157275












  • $begingroup$
    Thank you for answer.
    $endgroup$
    – Elementary
    Jan 5 at 14:03


















  • $begingroup$
    Thank you for answer.
    $endgroup$
    – Elementary
    Jan 5 at 14:03
















$begingroup$
Thank you for answer.
$endgroup$
– Elementary
Jan 5 at 14:03




$begingroup$
Thank you for answer.
$endgroup$
– Elementary
Jan 5 at 14:03


















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