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Atkinsoha

PreCalc & Trig Question! (Picture below) please try to be as specific as possible.. as I have not been able to go to school for over a week and have missed all of these lessons!

  • one year ago
  • one year ago

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  1. Atkinsoha
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    • one year ago
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  2. bamsanks
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    Try using the identity \[\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)\]Then \[\sin^2(x)+\cos^2(x)=1\] See how far you get with that.

    • one year ago
  3. Atkinsoha
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    I'm still lost.. how do you get from one to the other?

    • one year ago
  4. bamsanks
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    Right, if A = arcsin(x) and B = arccos(x), what would the addition identity give?

    • one year ago
  5. Atkinsoha
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    =arcsin(x) arccos(x) + no idea because sin is not given for b?

    • one year ago
  6. Atkinsoha
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    nor cos for A

    • one year ago
  7. bamsanks
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    Huh? Let: \[A=\sin^{-1}(x) \textrm{ and } B=\cos^{-1}(x)\]Then using the sum identity that I wrote: \[\sin(\sin^{-1}(x) + \cos^{-1}(x)) = \sin(\sin^{-1}(x))\cos(\cos^{-1}(x))+\sin(\cos^{-1}(x))\cos(\sin^{-1}(x))\]Follow up to here?

    • one year ago
  8. Atkinsoha
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    Kind of, yes.

    • one year ago
  9. bamsanks
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    Ok.. From here, we know that sin(arcsin(x)) = x and cos(arccos(x)) = x, so: \[ = x^2 + \sin(\cos^{-1}(x))\cos(\sin^{-1}(x))\] Now you can write \[\sin(x)=\sqrt{1-\cos^2(x)}\]and\[\cos(x)=\sqrt{1-\sin^2(x)}\]Using the second identity that I wrote. Again, do you follow?

    • one year ago
  10. Atkinsoha
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    okay, yes following so far.

    • one year ago
  11. bamsanks
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    Ok, so we now get \[=x^2 + \sqrt{1-\cos^2(\cos^{-1}(x))}\sqrt{1-\sin^2(\sin^{-1}(x))}\]\[=x^2 + \sqrt{1-x^2}\sqrt{1-x^2}\] Hopefully you can do the rest.. If there's anything you don't get, say so.

    • one year ago
  12. Atkinsoha
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    so.. it ends up being? (√1)?

    • one year ago
  13. bamsanks
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    Which simplifies to?

    • one year ago
  14. Atkinsoha
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    1..?

    • one year ago
  15. bamsanks
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    Exactly!

    • one year ago
  16. Atkinsoha
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    !!?! I get it! Thank you so much!

    • one year ago
  17. bamsanks
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    You're welcome.

    • one year ago
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