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anonymous

  • 5 years ago

Prove that: cos(sin^-1(x)) = sqrt (1-x^2)

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  1. watchmath
    • 5 years ago
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    Let \(a=\cos(\sin^{-1} x)\) Then \(a^2+\sin^2(\sin^{-1}x)=1\) But remember \(\sin(\sin^{-1}(x))=x\) Hence \(a^2+x^2=1\) \(a^2=1-x^2\) \(a=\sqrt{1-x^2}\) Therefore \(\cos(\sin^{-1}(x)=\sqrt{1-a^2}\)

  2. watchmath
    • 5 years ago
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    I mean \(\cos(\sin^{-1}(x))=\sqrt{1-x^2}\)

  3. anonymous
    • 5 years ago
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    Do you mind if I ask you another question

  4. watchmath
    • 5 years ago
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    If you think the answer above helps you please the good answer button :). Post your other question as a new question. If it is interesting, I might help you. But don't worry there are a lot of people who will be able to help you :D.

  5. anonymous
    • 5 years ago
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    hello wathmath!

  6. anonymous
    • 5 years ago
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    lots of posts tonight

  7. watchmath
    • 5 years ago
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    wow you already halfway amistre64 medals :D

  8. watchmath
    • 5 years ago
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    I don't know I am not as excited as before in answering questions :D

  9. anonymous
    • 5 years ago
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    well i have to agree. getting old, but my latex is improving by leaps and bounds

  10. anonymous
    • 5 years ago
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    and no amistre is way ahead. i will never catch up

  11. watchmath
    • 5 years ago
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    use \ ( \ ) instead of \ [ \ ] to get inline latex, that will make your post more concise :)

  12. anonymous
    • 5 years ago
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    yo satellite can you go back to this question that you were helping me with? If sin(x) = 1/3 and sec(y) = 5/4 , where x and y lie between 0 and pie/2, evaluate the expression cos(x+y).

  13. anonymous
    • 5 years ago
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    I asked you about how the pathagoras part you did i was a little confused..can you show that work?

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