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anonymous

  • one year ago

What is lim (cos((pi/2)+h) - cos(pi/2))/h? h->0

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  1. ganeshie8
    • one year ago
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    Hint : Let \(f(x)=\cos(x)\) the given expression is equivalent to \(f'(\pi/2)\)

  2. anonymous
    • one year ago
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    \[\frac{ \sin(h)-0}{ h }\]

  3. anonymous
    • one year ago
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    \[\frac{ \sin(h) }{ h }\]

  4. ganeshie8
    • one year ago
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    \[\large \frac{ \color{red}{-}\sin(h)-0}{ h }\] right ?

  5. ganeshie8
    • one year ago
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    because \(\cos(\pi/2 + x) = - \sin(x)\)

  6. anonymous
    • one year ago
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    ohhh ok! Is that like a identity?

  7. ganeshie8
    • one year ago
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    do you remember angle sum identity ? \(\cos(A+B)\)

  8. anonymous
    • one year ago
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    sort of, it sounds familiar

  9. ganeshie8
    • one year ago
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    \(\cos(A+B) = \cos A\cos B - \sin A\sin B\)

  10. ganeshie8
    • one year ago
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    use that to work out \(\cos(\pi/2+h)\)

  11. anonymous
    • one year ago
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    ohhhhhhh that makes so much sense now

  12. anonymous
    • one year ago
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    \[\frac{ -\sin(h) }{ h }\]

  13. anonymous
    • one year ago
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    So do I substitute 0 for h now?

  14. ganeshie8
    • one year ago
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    substitute h=0 and see what you get

  15. anonymous
    • one year ago
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    Oh yea that doesn't work with 0 on the bottom oops!

  16. ganeshie8
    • one year ago
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    right, try using the standard limits |dw:1440215549795:dw|

  17. anonymous
    • one year ago
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    So -sin(h)/h =-1?

  18. ganeshie8
    • one year ago
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    Yes : \[\large \lim\limits_{h\to 0}~\dfrac{ -\sin(h) }{ h } =- \lim\limits_{h\to 0}~\dfrac{ \sin(h) }{ h } =-1 \]

  19. anonymous
    • one year ago
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    Ohh I understand it all now!!! Thank you so much for teaching me you're great! :)

  20. ganeshie8
    • one year ago
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    np :)

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