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jayshane

  • 2 years 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
    • 2 years 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
    • 2 years ago
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    differentiate Y=cos4 t -sin4 t

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

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

  5. jayshane
    • 2 years ago
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    yes

  6. tkhunny
    • 2 years ago
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    What do you get for that?

  7. jayshane
    • 2 years ago
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    ist assingment dude

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

  9. jayshane
    • 2 years ago
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    its our assignment in differential calculus...

  10. zepdrix
    • 2 years 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
    • 2 years ago
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    the second one

  12. jayshane
    • 2 years ago
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    i dont know how to type that

  13. kirbykirby
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    the book has an answer of -2sin 2 t

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

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

  18. kirbykirby
    • 2 years 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
    • 2 years ago
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    Now use the double-angle formula

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

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

  22. jayshane
    • 2 years ago
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    THANK YOU KIRB!!!

  23. tkhunny
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    I'm usually the brute force guy. Once in a while I see one!

  27. kirbykirby
    • 2 years ago
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    Hehe good one ;)

  28. jayshane
    • 2 years ago
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    how about differentiate: y=sec²x-tan²x

  29. tkhunny
    • 2 years 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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