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Jonask

  • one year ago

trigonometry

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  1. Jonask
    • one year ago
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    \[\huge{\cos \frac{\pi}{7}}-\cos \frac{2\pi}{7}+\cos \frac{3\pi}{7}=\frac{1}{2}\]

  2. Jonask
    • one year ago
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    prove

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

  4. Jonask
    • one year ago
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    without using a calculator

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

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

  8. shubhamsrg
    • one year ago
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    *

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

  10. sauravshakya
    • one year ago
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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)

  11. shubhamsrg
    • one year ago
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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

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

  13. sauravshakya
    • one year ago
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    yes

  14. Jonask
    • one year ago
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    y

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

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

  17. Jonask
    • one year ago
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    k very clear

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

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

  20. Jonask
    • one year ago
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    on cos 5a

  21. Jonask
    • one year ago
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    right

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

  23. sauravshakya
    • one year ago
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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

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

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

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

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

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