How can I demonstrate that $cos(pi/8)=(sqrt2+sqrt2)/2$? [closed]

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I am "reminded" at the beginning of the exercise that $cosleft(dfracpi4right)=dfracsqrt22$ and that $cos (2x)=2cos^2(x)- 1$. I have been trying to figure this out:



$$cosdfracpi8 = dfracsqrt2 + sqrt22$$



for hours but keep going round in circles and it's driving me crazy.
It is probably extremely obvious to solve but I just can't seem to be able to.




trigonometry




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closed as off-topic by Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner Aug 23 at 8:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 6




    $$cos^2dfracpi8=dfrac1+cos(pi/4)2$$
    – Nosrati
    Aug 23 at 4:16






  • 1




    You say you’ve been going round in circles for hours, but you still need to include your attempts since this isn’t a site that will just give you answers.
    – Chase Ryan Taylor
    Aug 23 at 5:30














up vote
0
down vote

favorite












I am "reminded" at the beginning of the exercise that $cosleft(dfracpi4right)=dfracsqrt22$ and that $cos (2x)=2cos^2(x)- 1$. I have been trying to figure this out:



$$cosdfracpi8 = dfracsqrt2 + sqrt22$$



for hours but keep going round in circles and it's driving me crazy.
It is probably extremely obvious to solve but I just can't seem to be able to.




trigonometry




share|cite|improve this question














closed as off-topic by Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner Aug 23 at 8:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 6




    $$cos^2dfracpi8=dfrac1+cos(pi/4)2$$
    – Nosrati
    Aug 23 at 4:16






  • 1




    You say you’ve been going round in circles for hours, but you still need to include your attempts since this isn’t a site that will just give you answers.
    – Chase Ryan Taylor
    Aug 23 at 5:30












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am "reminded" at the beginning of the exercise that $cosleft(dfracpi4right)=dfracsqrt22$ and that $cos (2x)=2cos^2(x)- 1$. I have been trying to figure this out:



$$cosdfracpi8 = dfracsqrt2 + sqrt22$$



for hours but keep going round in circles and it's driving me crazy.
It is probably extremely obvious to solve but I just can't seem to be able to.




trigonometry




share|cite|improve this question














I am "reminded" at the beginning of the exercise that $cosleft(dfracpi4right)=dfracsqrt22$ and that $cos (2x)=2cos^2(x)- 1$. I have been trying to figure this out:



$$cosdfracpi8 = dfracsqrt2 + sqrt22$$



for hours but keep going round in circles and it's driving me crazy.
It is probably extremely obvious to solve but I just can't seem to be able to.





trigonometry





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 5:38









an4s

2,0632417




2,0632417










asked Aug 23 at 4:07









Carla Flores

81




81




closed as off-topic by Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner Aug 23 at 8:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner Aug 23 at 8:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Xander Henderson, Chase Ryan Taylor, mfl, Jendrik Stelzner
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 6




    $$cos^2dfracpi8=dfrac1+cos(pi/4)2$$
    – Nosrati
    Aug 23 at 4:16






  • 1




    You say you’ve been going round in circles for hours, but you still need to include your attempts since this isn’t a site that will just give you answers.
    – Chase Ryan Taylor
    Aug 23 at 5:30












  • 6




    $$cos^2dfracpi8=dfrac1+cos(pi/4)2$$
    – Nosrati
    Aug 23 at 4:16






  • 1




    You say you’ve been going round in circles for hours, but you still need to include your attempts since this isn’t a site that will just give you answers.
    – Chase Ryan Taylor
    Aug 23 at 5:30







6




6




$$cos^2dfracpi8=dfrac1+cos(pi/4)2$$
– Nosrati
Aug 23 at 4:16




$$cos^2dfracpi8=dfrac1+cos(pi/4)2$$
– Nosrati
Aug 23 at 4:16




1




1




You say you’ve been going round in circles for hours, but you still need to include your attempts since this isn’t a site that will just give you answers.
– Chase Ryan Taylor
Aug 23 at 5:30




You say you’ve been going round in circles for hours, but you still need to include your attempts since this isn’t a site that will just give you answers.
– Chase Ryan Taylor
Aug 23 at 5:30










1 Answer
1






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oldest

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up vote
4
down vote



accepted










Let $cosfracpi8 = c$



You know that $frac12 sqrt 2 = cos fracpi4 = 2c^2 - 1$



So you can set up the quadratic $2c^2 - 1 = frac 12sqrt 2$



$c^2 = frac 14 (2 + sqrt 2)$



$c = frac 12 sqrt2 +sqrt 2$ (taking the positive root based on the argument being in the first quadrant).






share|cite|improve this answer




















  • Thank you so so much!
    – Carla Flores
    Aug 23 at 5:01










  • @CarlaFlores Most welcome. :)
    – Deepak
    Aug 23 at 5:03

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Let $cosfracpi8 = c$



You know that $frac12 sqrt 2 = cos fracpi4 = 2c^2 - 1$



So you can set up the quadratic $2c^2 - 1 = frac 12sqrt 2$



$c^2 = frac 14 (2 + sqrt 2)$



$c = frac 12 sqrt2 +sqrt 2$ (taking the positive root based on the argument being in the first quadrant).






share|cite|improve this answer




















  • Thank you so so much!
    – Carla Flores
    Aug 23 at 5:01










  • @CarlaFlores Most welcome. :)
    – Deepak
    Aug 23 at 5:03














up vote
4
down vote



accepted










Let $cosfracpi8 = c$



You know that $frac12 sqrt 2 = cos fracpi4 = 2c^2 - 1$



So you can set up the quadratic $2c^2 - 1 = frac 12sqrt 2$



$c^2 = frac 14 (2 + sqrt 2)$



$c = frac 12 sqrt2 +sqrt 2$ (taking the positive root based on the argument being in the first quadrant).






share|cite|improve this answer




















  • Thank you so so much!
    – Carla Flores
    Aug 23 at 5:01










  • @CarlaFlores Most welcome. :)
    – Deepak
    Aug 23 at 5:03












up vote
4
down vote



accepted







up vote
4
down vote



accepted






Let $cosfracpi8 = c$



You know that $frac12 sqrt 2 = cos fracpi4 = 2c^2 - 1$



So you can set up the quadratic $2c^2 - 1 = frac 12sqrt 2$



$c^2 = frac 14 (2 + sqrt 2)$



$c = frac 12 sqrt2 +sqrt 2$ (taking the positive root based on the argument being in the first quadrant).






share|cite|improve this answer












Let $cosfracpi8 = c$



You know that $frac12 sqrt 2 = cos fracpi4 = 2c^2 - 1$



So you can set up the quadratic $2c^2 - 1 = frac 12sqrt 2$



$c^2 = frac 14 (2 + sqrt 2)$



$c = frac 12 sqrt2 +sqrt 2$ (taking the positive root based on the argument being in the first quadrant).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 23 at 4:14









Deepak

16.1k11437




16.1k11437











  • Thank you so so much!
    – Carla Flores
    Aug 23 at 5:01










  • @CarlaFlores Most welcome. :)
    – Deepak
    Aug 23 at 5:03
















  • Thank you so so much!
    – Carla Flores
    Aug 23 at 5:01










  • @CarlaFlores Most welcome. :)
    – Deepak
    Aug 23 at 5:03















Thank you so so much!
– Carla Flores
Aug 23 at 5:01




Thank you so so much!
– Carla Flores
Aug 23 at 5:01












@CarlaFlores Most welcome. :)
– Deepak
Aug 23 at 5:03




@CarlaFlores Most welcome. :)
– Deepak
Aug 23 at 5:03


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