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anonymous

  • one year ago

prove the following trigonometric identities. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x) tan(x/2)=sin(x)/(1+cos(x))

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  1. anonymous
    • one year ago
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    @dinamix

  2. anonymous
    • one year ago
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    \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x) \tan(x/2)=\frac{ \sin (x) }{1+\cos (x) } \] is that your question?

  3. anonymous
    • one year ago
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    no they are two separate equations. 2/(rt(3)cos(x)+sin(x))=sec(pi/6-x) tan(x/2)=sin(x)/(1+cos(x))

  4. anonymous
    • one year ago
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    oK

  5. anonymous
    • one year ago
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    I will do only one, which one do you like?

  6. anonymous
    • one year ago
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    do the top one

  7. anonymous
    • one year ago
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    you can do either one if you want

  8. IrishBoy123
    • one year ago
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    for the second \[tan\frac{x}{2}=\frac{sinx}{1+cosx}\] try this \[tan\frac{x}{2}=\frac{2sin\frac{x}{2}cos \frac{x}{2}}{1+(1-2sin^2 \frac{x}{2})}\]

  9. welshfella
    • one year ago
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    sec (pi/6 - x) = 1 / (sqrt3/2 cos x + sin pi/6 sin x) = 2 / (sqrt 3 cos + sin x)

  10. welshfella
    • one year ago
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    * I missed out the second step which is = 1 / (cos pi/6 cos x + sin pi/6 sin x)

  11. anonymous
    • one year ago
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    \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\sec(\pi/6-x)\] \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6-x)} }\] using:cos(u-v)=cos u cos v+sin u sin v \[\frac{ 2 }{\sqrt{3} \cos(x)+\sin(x)}=\frac{ 1 }{ \cos{(\pi/6) \times \cos(x)}+\sin{(\pi/6) \times \sin(x)} }\]

  12. anonymous
    • one year ago
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    i got that far ASAAD123 put i didn't know what to do after that

  13. welshfella
    • one year ago
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    the denominator = cos (pi/6 - x)

  14. welshfella
    • one year ago
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    and 1 / cos = sec

  15. anonymous
    • one year ago
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    wouldn't you also change sin(pi/6) to 1/2

  16. welshfella
    • one year ago
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    no its the sin of a compound angle (pi/6 - x) not sin pi/6

  17. anonymous
    • one year ago
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    ok

  18. anonymous
    • one year ago
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  19. anonymous
    • one year ago
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    so how do you get the sides to equal

  20. welshfella
    • one year ago
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    i had to use sin pi/6 = 1/2 because i started with RHS and converted to LHS whereas ASAAD did the reverse

  21. welshfella
    • one year ago
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    with these identities you choose one side and try to convert it to the other. This proves the identity.

  22. anonymous
    • one year ago
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    ok

  23. anonymous
    • one year ago
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    i think i understand it now

  24. welshfella
    • one year ago
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    so you can use asaad's or mine. Either would do.

  25. welshfella
    • one year ago
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    There is only more more step to prove IrishBoy's solution

  26. anonymous
    • one year ago
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    i already got irish boy's solution

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