Prove that cos^2 3π/7 - cos^2 3π/14 + cos^2 π/7 = 1/4

[achyuth]

Have been trying this for 2 hours and not getting. I went till making the equation cos4π/7cos2π/7 + sin^2 3π/14. Further simplification is making it more complicated.

 

[Deleted member 4993]

Have been trying this for 2 hours and not getting. I went till making the equation cos4π/7cos2π/7 + sin^2 3π/14. Further simplification is making it more complicated.

cos^2 3π/7 - cos^2 3π/14 + cos^2 π/7 = 1/4

substitute

u = π/14................... to get

cos²(6u) - cos²(3u) + cos²(2u) = 1/4

Next I would use:

cos²(t) = [cos(2t) +1]/2 ........................... to eliminate those pesky 'squares'.

Please show us what you have tried and exactly where you are stuck.

Please share your work/thoughts about this problem.

 

[Lindsay Hotham]

cos^2 3π/7 - cos^2 3π/14 + cos^2 π/7 = 1/4

substitute

u = π/14................... to get

cos²(6u) - cos²(3u) + cos²(2u) = 1/4

Next I would use:

cos²(t) = [cos(2t) +1]/2 ........................... to eliminate those pesky 'squares'.

Please show us what you have tried and exactly where you are stuck.

Please share your work/thoughts about this problem.

I don't think what you are suggesting will work. Because 7 is a prime number any of the usual cosine or sine transformations will simply lead to further angles with denominators of 7 14 or 28.

 

[Deleted member 4993]

I don't think what you are suggesting will work. Because 7 is a prime number any of the usual cosine or sine transformations will simply lead to further angles with denominators of 7 14 or 28.

What would be your proposed method?