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anonymous

  • one year ago

find the exact value of cos(sin^-1(3/5))

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  1. anonymous
    • one year ago
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    Do you have a calculator?

  2. anonymous
    • one year ago
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    I got \( \huge \frac{4}{5} \)

  3. Michele_Laino
    • one year ago
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    If \theta is such that: \[\Large \theta = \arcsin \left( {\frac{3}{5}} \right)\] then you have to compute this: \[\Large \cos \theta \]

  4. anonymous
    • one year ago
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    were not suppose to use calculator :/

  5. anonymous
    • one year ago
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    whats arc sin?

  6. Michele_Laino
    • one year ago
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    ok! now we have: \[\Large \sin \theta = \frac{3}{5}\] am I right?

  7. Michele_Laino
    • one year ago
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    it is another way to mean sin^-1

  8. Michele_Laino
    • one year ago
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    so, if we apply the fundamental identity: \[\Large {\left( {\cos \theta } \right)^2} + {\left( {\sin \theta } \right)^2} = 1\] we can write: \[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = ...?\]

  9. anonymous
    • one year ago
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    let 2 sin^-1(3/5) = A sin^-1(3/5) = A/2 sin(A/2) = 3/5 cos A = 1 - 2sin^2(A/2) = 1 - 2(9/25) = 7/25 Thus cos [ 2sin^-1(3/5)] = cos A = 7/25 I found this answer but where could they have gotten 9 from?

  10. Michele_Laino
    • one year ago
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    no, we have: cos(A/2) =sqrt(1 - sin^2(A/2)

  11. anonymous
    • one year ago
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    wait so the way they did it is wrong?

  12. Michele_Laino
    • one year ago
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    yes! I think so!

  13. Michele_Laino
    • one year ago
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    it is not necessary to work with A/2. Please apply my procedure

  14. Michele_Laino
    • one year ago
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    I used +/- since we have two square roots

  15. Michele_Laino
    • one year ago
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    for example: \[\Large \sqrt 4 = \pm 2\]

  16. anonymous
    • one year ago
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    Wait so how would your work look like then?

  17. anonymous
    • one year ago
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    Because your procedure is confusing me even more lol

  18. Michele_Laino
    • one year ago
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    since we have: \[\Large \sin \theta = \frac{3}{5}\] then we can write: \[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = ...?\]

  19. anonymous
    • one year ago
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    is this a half angle formula?

  20. Michele_Laino
    • one year ago
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    no, it is not a bisection formula

  21. Michele_Laino
    • one year ago
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    hint: \[\Large \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{25 - 9}}{{25}}} = ...?\]

  22. anonymous
    • one year ago
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    Is there another way to do it? becauase our teacher didn't show us that formula.

  23. Michele_Laino
    • one year ago
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    Sincerely speaking, it is the unique way that I know

  24. anonymous
    • one year ago
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    but whats wrong with the method i used? im still not understanding

  25. Michele_Laino
    • one year ago
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    your method is correct, nevertheless you have to find cos(A/2) not cos(A)

  26. Michele_Laino
    • one year ago
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    you have used the subsequent bisection formula: \[\Large {\left\{ {\sin \left( {\frac{\theta }{2}} \right)} \right\}^2} = \frac{{1 - \cos \theta }}{2}\]

  27. anonymous
    • one year ago
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    This question just got worse so I used something else but idk if its riight..

  28. Michele_Laino
    • one year ago
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    the computation of your image is correct

  29. anonymous
    • one year ago
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    So is 7/25 right then??

  30. anonymous
    • one year ago
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    @Michele_Laino

  31. Michele_Laino
    • one year ago
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    we can write this: \[\Large \sin \theta = \frac{3}{5} \Rightarrow \cos \left( {2\theta } \right) = \frac{7}{{25}}\]

  32. anonymous
    • one year ago
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    @Michele_Laino lol what does that mean tho? XD can you explain it?

  33. Michele_Laino
    • one year ago
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    it is simple: since \[\Large \sin \theta = \frac{3}{5}\] then: \[\Large \cos \theta = \pm \sqrt {1 - {{\left( {\sin \theta } \right)}^2}} = \pm \sqrt {1 - {{\left( {\frac{3}{5}} \right)}^2}} = \pm \frac{4}{5}\] Now we use the duplication formula: \[\Large \begin{gathered} \cos \left( {2\theta } \right) = {\left( {\cos \theta } \right)^2} - {\left( {\sin \theta } \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{{16 - 9}}{{25}} = \frac{7}{{25}} \hfill \\ \end{gathered} \]

  34. anonymous
    • one year ago
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    ohh alright its the double angle formula too!

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