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owlet

  • one year ago

Help me solve this please: \(\sf sin(cos^{-1} (-3/5))\)

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  1. FibonacciChick666
    • one year ago
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    So let's start with what will you do first?

  2. FibonacciChick666
    • one year ago
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    (by order of ops)

  3. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sin(\cos^{-1}x)=\sqrt{1-\left[\cos(\cos^{-1}x)\right]^2~~} }\)

  4. SolomonZelman
    • one year ago
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    using, the fact that: sin²(w) + cos²(w) = 1 sin²(w) = 1 - cos²(w) sin(w) = \(\sqrt{1 - \cos^2(w)}\) in this case w is: \(\cos^{-1}(-3/5)\) (That makes it even easier for you)

  5. SolomonZelman
    • one year ago
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    And also, you should know that: \(\large\color{black}{ \displaystyle \left[\cos(\cos^{-1}\theta)\right]=\theta }\)

  6. owlet
    • one year ago
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    wait i'm confused.. why did you take the sqrt ?

  7. SolomonZelman
    • one year ago
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    because I had sin²(quantity) and I want to solve for just sin(quantity)

  8. Loser66
    • one year ago
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    Support @SolomonZelman \(\pm \sqrt...\)

  9. FibonacciChick666
    • one year ago
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    @SolomonZelman nice use of identity, but it's a special triangle. If one recognizes that it can be done in two steps

  10. SolomonZelman
    • one year ago
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    Well, it is a number in our case, and I was assuming that just like by taking a square root of any ordinary number such as √35 you don't have a ±, so it would be here without a ±....

  11. SolomonZelman
    • one year ago
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    (Well, it is a single value expression, isn't it ? )

  12. owlet
    • one year ago
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    @FibonacciChick666 my first attempt, I also used special triangles and the answer that I got is -4/5 but it's not right

  13. Loser66
    • one year ago
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    cos (theta) = negative number, then theta can be in Quadrant 2 or Quadrant 3. That is why when taking off the square root, you need +/-

  14. owlet
    • one year ago
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    okay, I understand that part, but how am I going to apply it to solve for cos^-1 (-3/5)

  15. Loser66
    • one year ago
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    \(cos (\alpha) = A \) that is \(cos^{-1} A = \alpha\), right?

  16. Loser66
    • one year ago
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    Hence \(cos (\alpha) = -3/5= \dfrac{adj}{hypotenuse}\)

  17. SolomonZelman
    • one year ago
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    owlet your answer, I think that should be positive.

  18. Loser66
    • one year ago
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    |dw:1443625732989:dw|

  19. Loser66
    • one year ago
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    now, \(sin(\alpha) = \dfrac{opposite}{hypotenuse}= 4/5\)

  20. Loser66
    • one year ago
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    However, when alpha is in Quadrant 3, you still have cos alpha = -3/5

  21. FibonacciChick666
    • one year ago
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    What loser is saying is what I was getting at

  22. Loser66
    • one year ago
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    |dw:1443625863982:dw|

  23. Loser66
    • one year ago
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    hence sin (alpha) = -4/5 also

  24. Loser66
    • one year ago
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    Your answer should be both. I think!!

  25. FibonacciChick666
    • one year ago
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    I agree with Loser given no stipulations on quadrant

  26. SolomonZelman
    • one year ago
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    i checked wolfram gives only 4/5 positive. http://www.wolframalpha.com/input/?i=sin%28cos%5E%28-1%29+of+%28-3%2F5%29%29

  27. SolomonZelman
    • one year ago
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    So it assumes that it is in the 2nd quadrant.

  28. Loser66
    • one year ago
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    @SolomonZelman It's nice but to us (@FibonacciChick666 and me) we need the firm logic to make sure that we don't miss any cases on our argument. :)

  29. SolomonZelman
    • one year ago
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    Sure, I am not against -4/5:)

  30. owlet
    • one year ago
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    i was doing the same thing as loser.. but yeah i think that the right answer is positive 4/5 but idk why the right is only positive

  31. Loser66
    • one year ago
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    You may have some restriction on the problem, like the angle is in Q2. or (0, pi) for domain.

  32. owlet
    • one year ago
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    okay. thanks :)

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