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












5














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




















  • 1




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












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










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
















5














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




















  • 1




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












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










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














5












5








5


0





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















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

47.7k870151




47.7k870151










asked Jan 4 at 11:22









BeginnerBeginner

30510




30510








  • 1




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












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










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














  • 1




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












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










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








1




1




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






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














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




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












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




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










2 Answers
2






active

oldest

votes


















3














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























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










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






  • 1




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



















1














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





















  • Thank you for answer.
    – Beginner
    Jan 5 at 14:03











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

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active

oldest

votes









3














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























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










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






  • 1




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
















3














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























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










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






  • 1




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














3












3








3






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














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

7,8081037




7,8081037












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










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






  • 1




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


















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










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






  • 1




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
















(+1) Many thanks. I hope we can complete.
– Beginner
Jan 4 at 12:19




(+1) Many thanks. I hope we can complete.
– Beginner
Jan 4 at 12:19












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




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




1




1




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




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











1














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





















  • Thank you for answer.
    – Beginner
    Jan 5 at 14:03
















1














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





















  • Thank you for answer.
    – Beginner
    Jan 5 at 14:03














1












1








1






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












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

224k15156274




224k15156274












  • Thank you for answer.
    – Beginner
    Jan 5 at 14:03


















  • Thank you for answer.
    – Beginner
    Jan 5 at 14:03
















Thank you for answer.
– Beginner
Jan 5 at 14:03




Thank you for answer.
– Beginner
Jan 5 at 14:03


















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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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1300-talet

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Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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Display a custom attribute below product name in the front-end Magento 1.9.3.8

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0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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