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rudra333

  • 3 years ago

arccos0?

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  1. tomo
    • 3 years ago
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    1/cos(0)

  2. rudra333
    • 3 years ago
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    \[\tan ^{-1}-\frac{ \sqrt{3} }{ 3}\]

  3. rudra333
    • 3 years ago
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    @tomo is that the asnwer?

  4. tomo
    • 3 years ago
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    ?? that doesn't make sense. please reword

  5. rudra333
    • 3 years ago
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    the one with tangent?

  6. rudra333
    • 3 years ago
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    what is arccos0??

  7. tomo
    • 3 years ago
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    cos(0) = 1.

  8. rudra333
    • 3 years ago
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    @tomo

  9. Inspired
    • 3 years ago
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    arccos(0)= 90 degrees or pi/2

  10. tomo
    • 3 years ago
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    one second. it is a little different than i first thought it was.

  11. rudra333
    • 3 years ago
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    k but plz try to expalin i have test on this toorrow

  12. rudra333
    • 3 years ago
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    k what is the answer of the questio with tangent?

  13. rudra333
    • 3 years ago
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    @Inspired

  14. Inspired
    • 3 years ago
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    Well it's just the inverse of cos(0). Cos(0) =1. The inverse of that switches the values of the sine cosine functions. cos(90) gives you 1. but because remember, sin(0)=0 and sin(90)=1

  15. Inspired
    • 3 years ago
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    cos(90) is 0*

  16. Inspired
    • 3 years ago
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    and an easy way to remember is: sin is your y axis and cos is your x axis. the inverse of a function is basically switches the x and y. what was y=f(x) is now x=f(y)

  17. rudra333
    • 3 years ago
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    k i know that but i have like 5 problems to work on i just want the right answers to it so i can practice

  18. rudra333
    • 3 years ago
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    for the test tomorrow

  19. rudra333
    • 3 years ago
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    socan u tell me the answer to the tangent one?plz

  20. rudra333
    • 3 years ago
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    \[\arcsin-\frac{ \sqrt{3} }{ 2 }\]

  21. rudra333
    • 3 years ago
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    and this one plz i have to go emergency plz answe those 2 for me plz

  22. rudra333
    • 3 years ago
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    that will be really helpful

  23. Inspired
    • 3 years ago
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    -pi/6 and -pi/3 (take a look at the unit circle for the second one)

  24. fgoh
    • 3 years ago
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    how did u get -pi/6? i know in the unit cirlce but when i tried all of them tan is y/x then which onw will work?

  25. fgoh
    • 3 years ago
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    @Inspired

  26. Inspired
    • 3 years ago
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    Let's see. arctan (sqrt3/3) is the same thing as arctan (1/sqrt3) right? and using 30-60-90 triangle, you have this: |dw:1358819026253:dw|

  27. Inspired
    • 3 years ago
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    Anyway, since the value is negative, it will occur in the 2nd or 4th Quadrant. Since we want a single value, you usually use the 4th quadrant, and that gives you a negative value.

  28. fgoh
    • 3 years ago
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    but is is y/x right why it doesn't work

  29. fgoh
    • 3 years ago
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    if u do it with the values on unit square and i thought the domain is from (pi/2,-pi/2)?

  30. fgoh
    • 3 years ago
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    |dw:1358819636475:dw|

  31. fgoh
    • 3 years ago
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    OMG K GET IT THANKS THANKS ALOT @Inspired U ARE AWESOME AND HELPFUL OMG THANKS THANKS I JUST FEEL SO HAPPY WHEN I GET IT THANKS

  32. Inspired
    • 3 years ago
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    :)

  33. fgoh
    • 3 years ago
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    \[COS ^{-1}(\sin 11\pi/6)\]

  34. fgoh
    • 3 years ago
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    @Inspired

  35. fgoh
    • 3 years ago
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    @Inspired last one

  36. fgoh
    • 3 years ago
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    and which one do u do first?

  37. Inspired
    • 3 years ago
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    ok :0 sin(11pi/6) is the same thing as sin(12pi/6-pi/6) so you get sin(2pi-pi/6) --> -sin(pi6)=-1/2 and so you have arccos(-1/2)

  38. rudra333
    • 3 years ago
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    is that the answer arccos(-1/20)

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