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jayshane

  • one year ago

DERIVATIVES OF TRIGO IDENTITIES no.1)Y=cos4 t- sin4 t=-2sin 2t INEED SOLUTION PLS....T_T NID BADLY!!!!

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  1. tkhunny
    • one year ago
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    It makes no sense as it it presented. Trigonometric Identities are statements and the term "derivative" has no meaning for such a statement. Can you provide the exact wording of the problem?

  2. jayshane
    • one year ago
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    differentiate Y=cos4 t -sin4 t

  3. kirbykirby
    • one year ago
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    Is this cos(4t) or \[\cos^4t\]

  4. tkhunny
    • one year ago
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    We're looking for \(\dfrac{dY}{dt}\)? Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)

  5. jayshane
    • one year ago
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    yes

  6. tkhunny
    • one year ago
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    What do you get for that?

  7. jayshane
    • one year ago
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    ist assingment dude

  8. tkhunny
    • one year ago
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    Conversation Overlap Misunderstanding... What do you get for \(\dfrac{d}{dt}\cos^{4}(t)\)

  9. jayshane
    • one year ago
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    its our assignment in differential calculus...

  10. zepdrix
    • one year ago
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    Jay you didn't answer the question kirby asked. Is it suppose to be \(\cos(4t)\) or \(\cos^4(t)\)?

  11. jayshane
    • one year ago
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    the second one

  12. jayshane
    • one year ago
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    i dont know how to type that

  13. kirbykirby
    • one year ago
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    Just think of it as the chain rule. You can move the exponent if it confuses you: \[\frac{d}{dt}\cos^4t=\frac{d}{dt}(\cos t)^4 = 4(\cos t)^3(-\sin t)\]

  14. kirbykirby
    • one year ago
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    -sin t comes from the fact that it is the derivative of cos t by using the chain rule.

  15. jayshane
    • one year ago
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    the book has an answer of -2sin 2 t

  16. kirbykirby
    • one year ago
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    Just use the same logic on \[\frac{d}{dt}\sin^4t=4\sin^3t(\cos t)\]

  17. jayshane
    • one year ago
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    cos4t-sin4t= kirb can u give me the whole solution soo i can studied it... please

  18. kirbykirby
    • one year ago
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    so now: \[-4\cos^3t(\sin t) - 4\sin^3t(\cos t) = -4\cos t*\sin t(\cos^t+\sin^2t)=-4\cos t*\sin t(1) \]

  19. kirbykirby
    • one year ago
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    Now use the double-angle formula

  20. kirbykirby
    • one year ago
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    \[Since : \sin(2t)=2\sin t \cos t\]

  21. kirbykirby
    • one year ago
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    \[-4\cos t \sin t = -2(2\cos t \sin t) = -2\sin(2t)\]

  22. jayshane
    • one year ago
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    THANK YOU KIRB!!!

  23. tkhunny
    • one year ago
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    Since we're just doing your homework, I'd do it this way. \(\cos^{4}(t) - \sin^{4}(t) = [\cos^{2}(t) - \sin^{2}(t)][\cos^{2}(t) + \sin^{2}(t)] = \cos(2x)\) It's a lot easier after that.

  24. tkhunny
    • one year ago
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    Sorry, not sure why I wrote 2x on the end, there. Should be 2t. If you are going to make me do ALL the work, you'll have to show me how that last step happened. There's a whole lot of stuff in there that magically turned into cos(2t).

  25. kirbykirby
    • one year ago
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    The method by tkhunny is also excellent :) It is shorter but I usually just do it "straight-forward" unless I'm stuck and tkhunny's method is a good trick to make the derivative a lot easier

  26. tkhunny
    • one year ago
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    I'm usually the brute force guy. Once in a while I see one!

  27. kirbykirby
    • one year ago
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    Hehe good one ;)

  28. jayshane
    • one year ago
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    how about differentiate: y=sec²x-tan²x

  29. tkhunny
    • one year ago
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    Go right ahead. Let's see your first attempt. Hint: It is amazingly trivial! Remember your trigonometry. This is not much of a calculus problem.

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