Mixing
If $sin 5°+sin 10°+sin15°+cdots+sin 40°=a$, then $sin 5°+sin 10°+sin15°+cdots+sin 175°=?$
$begingroup$
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
edited Jan 4 at 11:29
Blue
47.8k870152
asked Jan 4 at 11:22
BeginnerBeginner
34110
$endgroup$
add a comment |
$begingroup$
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
edited Jan 4 at 11:29
Blue
47.8k870152
asked Jan 4 at 11:22
BeginnerBeginner
34110
$endgroup$
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
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You can link to your question more easier than anybody.
$endgroup$
– kelalaka
Jan 4 at 15:02
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math.stackexchange.com/questions/17966/…
$endgroup$
– lab bhattacharjee
Jan 4 at 19:47
add a comment |
$begingroup$
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
edited Jan 4 at 11:29
Blue
47.8k870152
asked Jan 4 at 11:22
BeginnerBeginner
34110
$endgroup$
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
edited Jan 4 at 11:29
Blue
47.8k870152
asked Jan 4 at 11:22
BeginnerBeginner
34110
edited Jan 4 at 11:29
Blue
47.8k870152
asked Jan 4 at 11:22
BeginnerBeginner
34110
edited Jan 4 at 11:29
Blue
47.8k870152
edited Jan 4 at 11:29
Blue
47.8k870152
edited Jan 4 at 11:29
Blue
47.8k870152
47.8k870152
asked Jan 4 at 11:22
BeginnerBeginner
34110
asked Jan 4 at 11:22
BeginnerBeginner
34110
asked Jan 4 at 11:22
BeginnerBeginner
34110
34110
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
$endgroup$
$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Beginner
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
add a comment |
$begingroup$
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 $sin2.5^circcdotcos2.5^circne0$
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Thank you for answer.
$endgroup$
– Beginner
Jan 5 at 14:03
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
$endgroup$
$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Beginner
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
add a comment |
$begingroup$
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 $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
$endgroup$
$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Beginner
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
add a comment |
$begingroup$
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 $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
$endgroup$
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 $b = frac{left( sqrt{2}+2right) , left( 4 a+sqrt{2}right) }{2}$
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
edited Jan 4 at 13:32
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
answered Jan 4 at 12:05
AndreasAndreas
7,8921037
7,8921037
$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Beginner
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
add a comment |
$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Beginner
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.
$endgroup$
– Beginner
Jan 4 at 12:19
$begingroup$
(+1) Many thanks. I hope we can complete.
$endgroup$
– Beginner
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
add a comment |
$begingroup$
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 $sin2.5^circcdotcos2.5^circne0$
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Thank you for answer.
$endgroup$
– Beginner
Jan 5 at 14:03
add a comment |
$begingroup$
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 $sin2.5^circcdotcos2.5^circne0$
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Thank you for answer.
$endgroup$
– Beginner
Jan 5 at 14:03
add a comment |
$begingroup$
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 $sin2.5^circcdotcos2.5^circne0$
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
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 $sin2.5^circcdotcos2.5^circne0$
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 5 at 13:25
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
$begingroup$
Thank you for answer.
$endgroup$
– Beginner
Jan 5 at 14:03
add a comment |
$begingroup$
Thank you for answer.
$endgroup$
– Beginner
Jan 5 at 14:03
$begingroup$
Thank you for answer.
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– Beginner
Jan 5 at 14:03
$begingroup$
Thank you for answer.
$endgroup$
– Beginner
Jan 5 at 14:03
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$begingroup$
$sin(pi/4+x)=frac{1}{sqrt 2}(sin x+cos x)$,$sin(pi/2+x)=cos x$,$sin(3pi/4+x)=frac{1}{sqrt 2}(sin x-cos x)$-These identities may be helpful.
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– Thomas Shelby
Jan 4 at 12:20
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You can link to your question more easier than anybody.
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– kelalaka
Jan 4 at 15:02
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math.stackexchange.com/questions/17966/…
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– lab bhattacharjee
Jan 4 at 19:47