anonymous
  • anonymous
DERIVATIVES OF TRIGO IDENTITIES no.1)Y=cos4 t- sin4 t=-2sin 2t INEED SOLUTION PLS....T_T NID BADLY!!!!
Mathematics
chestercat
  • chestercat
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tkhunny
  • tkhunny
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?
anonymous
  • anonymous
differentiate Y=cos4 t -sin4 t
kirbykirby
  • kirbykirby
Is this cos(4t) or \[\cos^4t\]

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tkhunny
  • tkhunny
We're looking for \(\dfrac{dY}{dt}\)? Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)
anonymous
  • anonymous
yes
tkhunny
  • tkhunny
What do you get for that?
anonymous
  • anonymous
ist assingment dude
tkhunny
  • tkhunny
Conversation Overlap Misunderstanding... What do you get for \(\dfrac{d}{dt}\cos^{4}(t)\)
anonymous
  • anonymous
its our assignment in differential calculus...
zepdrix
  • zepdrix
Jay you didn't answer the question kirby asked. Is it suppose to be \(\cos(4t)\) or \(\cos^4(t)\)?
anonymous
  • anonymous
the second one
anonymous
  • anonymous
i dont know how to type that
kirbykirby
  • kirbykirby
Just think of it as the chain rule. You can move the exponent if it confuses you: \[\frac{d}{dt}\cos^4t=\frac{d}{dt}(\cos t)^4 = 4(\cos t)^3(-\sin t)\]
kirbykirby
  • kirbykirby
-sin t comes from the fact that it is the derivative of cos t by using the chain rule.
anonymous
  • anonymous
the book has an answer of -2sin 2 t
kirbykirby
  • kirbykirby
Just use the same logic on \[\frac{d}{dt}\sin^4t=4\sin^3t(\cos t)\]
anonymous
  • anonymous
cos4t-sin4t= kirb can u give me the whole solution soo i can studied it... please
kirbykirby
  • kirbykirby
so now: \[-4\cos^3t(\sin t) - 4\sin^3t(\cos t) = -4\cos t*\sin t(\cos^t+\sin^2t)=-4\cos t*\sin t(1) \]
kirbykirby
  • kirbykirby
Now use the double-angle formula
kirbykirby
  • kirbykirby
\[Since : \sin(2t)=2\sin t \cos t\]
kirbykirby
  • kirbykirby
\[-4\cos t \sin t = -2(2\cos t \sin t) = -2\sin(2t)\]
anonymous
  • anonymous
THANK YOU KIRB!!!
tkhunny
  • tkhunny
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.
tkhunny
  • tkhunny
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).
kirbykirby
  • kirbykirby
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
tkhunny
  • tkhunny
I'm usually the brute force guy. Once in a while I see one!
kirbykirby
  • kirbykirby
Hehe good one ;)
anonymous
  • anonymous
how about differentiate: y=sec²x-tan²x
tkhunny
  • tkhunny
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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