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 2 years ago
DERIVATIVES OF TRIGO IDENTITIES
no.1)Y=cos4 t sin4 t=2sin 2t INEED SOLUTION PLS....T_T NID BADLY!!!!
 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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tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1It 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?

jayshane
 2 years ago
Best ResponseYou've already chosen the best response.0differentiate Y=cos4 t sin4 t

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.3Is this cos(4t) or \[\cos^4t\]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1We're looking for \(\dfrac{dY}{dt}\)? Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1What do you get for that?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Conversation Overlap Misunderstanding... What do you get for \(\dfrac{d}{dt}\cos^{4}(t)\)

jayshane
 2 years ago
Best ResponseYou've already chosen the best response.0its our assignment in differential calculus...

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Jay you didn't answer the question kirby asked. Is it suppose to be \(\cos(4t)\) or \(\cos^4(t)\)?

jayshane
 2 years ago
Best ResponseYou've already chosen the best response.0i dont know how to type that

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.3Just 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
 2 years ago
Best ResponseYou've already chosen the best response.3sin t comes from the fact that it is the derivative of cos t by using the chain rule.

jayshane
 2 years ago
Best ResponseYou've already chosen the best response.0the book has an answer of 2sin 2 t

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.3Just use the same logic on \[\frac{d}{dt}\sin^4t=4\sin^3t(\cos t)\]

jayshane
 2 years ago
Best ResponseYou've already chosen the best response.0cos4tsin4t= kirb can u give me the whole solution soo i can studied it... please

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.3so 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
 2 years ago
Best ResponseYou've already chosen the best response.3Now use the doubleangle formula

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.3\[Since : \sin(2t)=2\sin t \cos t\]

kirbykirby
 2 years ago
Best ResponseYou've already chosen the best response.3\[4\cos t \sin t = 2(2\cos t \sin t) = 2\sin(2t)\]

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Since 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
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry, 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
 2 years ago
Best ResponseYou've already chosen the best response.3The method by tkhunny is also excellent :) It is shorter but I usually just do it "straightforward" unless I'm stuck and tkhunny's method is a good trick to make the derivative a lot easier

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1I'm usually the brute force guy. Once in a while I see one!

jayshane
 2 years ago
Best ResponseYou've already chosen the best response.0how about differentiate: y=sec²xtan²x

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Go 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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