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Jonask Group Title

trigonometry

  • one year ago
  • one year ago

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  1. Jonask Group Title
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    \[\huge{\cos \frac{\pi}{7}}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7}=\frac{1}{2}\]

    • one year ago
  2. Jonask Group Title
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    prove

    • one year ago
  3. Jonask Group Title
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    so far i managed to write \[\cos \frac{ \pi }{ 7 }-2\cos^2\frac{ \pi }{ 7 }+1+\cos \frac{ 2\pi }{ 7 }\cos \frac{ \pi }{ 7 }-\sin \frac{ 2\pi }{ 7 }\sin \frac{ \pi }{ 7 }\] \[\cos \frac{ \pi }{ 7 }-2\cos ^2\frac{ \pi }{ 7 }+1+2\cos^3\frac{ \pi }{ 7 }-\cos \frac{ \pi }{ 7 }-2\cos \frac{ \pi }{ 7 }+2\cos^3\frac{ \pi }{ 7 }\] \[4\cos^3\frac{ \pi }{ 7 }-2\cos^2\frac{ \pi }{ 7 }-2\cos\frac{\pi}{7}+1\]

    • one year ago
  4. Jonask Group Title
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    without using a calculator

    • one year ago
  5. Jonask Group Title
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    guys i have the solution but it is so superficial i'll put it just now

    • one year ago
  6. Jonask Group Title
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    • one year ago
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  7. Jonask Group Title
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    no idea about S

    • one year ago
  8. shubhamsrg Group Title
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    *

    • one year ago
  9. Jonask Group Title
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    is that the bookmark symbol

    • one year ago
  10. sauravshakya Group Title
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    cos(pi/7) -cos(2pi/7)+ cos(3pi/7) cos(pi/7)+cos(3pi/7)+cos(5pi/7) {2sin(pi/7)cos(pi/7) + 2 sin(pi/7) cos(3pi/7) + 2 sin(pi/7) cos(5pi/7)}/2 sin(pi/7)

    • one year ago
  11. shubhamsrg Group Title
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    That is just a comment, which will enable me to receive notifications whenever it gets discussed * is just a symbol, yes, bookmark symbol

    • one year ago
  12. Jonask Group Title
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    are you saying -cos 2 pi/7=cos 5 pi/7

    • one year ago
  13. sauravshakya Group Title
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    yes

    • one year ago
  14. Jonask Group Title
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    y

    • one year ago
  15. Jonask Group Title
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    -cos 2a=cos 5a

    • one year ago
  16. sauravshakya Group Title
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    cos(5pi/7) =cos(pi -2pi/7) = -cos(2pi/7)

    • one year ago
  17. Jonask Group Title
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    k very clear

    • one year ago
  18. sauravshakya Group Title
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    It is because cos(pi-a)=-cosa

    • one year ago
  19. sauravshakya Group Title
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    Now, u need to use 2 sinA cosB = sin(A+B) - sin(A-B)

    • one year ago
  20. Jonask Group Title
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    on cos 5a

    • one year ago
  21. Jonask Group Title
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    right

    • one year ago
  22. sauravshakya Group Title
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    Sorry, The correct formula is 2 sinA cosB =sin(A+B) + sin(A-B) =sin(A+B) - sin(B-A)

    • one year ago
  23. sauravshakya Group Title
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    =cos(pi/7) -cos(2pi/7)+ cos(3pi/7) =cos(pi/7)+cos(3pi/7)+cos(5pi/7) ={2sin(pi/7)cos(pi/7) + 2 sin(pi/7) cos(3pi/7) + 2 sin(pi/7) cos(5pi/7)}/2 sin(pi/7) ={sin(2pi/7) -sin(0) +sin(4pi/7) -sin(2pi/7) +sin(6pi/7) -sin(4pi/7)}/2sin(pi/7) ={sin(6pi/7)}/2sin(pi/7) ={sin(pi/7)}/2sin(pi/7) =1/2

    • one year ago
  24. sauravshakya Group Title
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    sin(6pi/7) =sin(pi -pi/7) =sin(pi/7)

    • one year ago
  25. Jonask Group Title
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    okay that makes so much sense i guess elementary steps we skipped

    • one year ago
  26. Jonask Group Title
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    so we just had to play with\[\pi \pm \frac{\pi}{7}\]

    • one year ago
  27. sauravshakya Group Title
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    yep... and one formula

    • one year ago
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