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

- anonymous

- chestercat

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- 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

differentiate Y=cos4 t -sin4 t

- kirbykirby

Is this cos(4t) or \[\cos^4t\]

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## More answers

- tkhunny

We're looking for \(\dfrac{dY}{dt}\)?
Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)

- anonymous

yes

- tkhunny

What do you get for that?

- anonymous

ist assingment dude

- tkhunny

Conversation Overlap Misunderstanding...
What do you get for \(\dfrac{d}{dt}\cos^{4}(t)\)

- anonymous

its our assignment in differential calculus...

- zepdrix

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

- anonymous

the second one

- anonymous

i dont know how to type that

- 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

-sin t comes from the fact that it is the derivative of cos t by using the chain rule.

- anonymous

the book has an answer of -2sin 2 t

- kirbykirby

Just use the same logic on \[\frac{d}{dt}\sin^4t=4\sin^3t(\cos t)\]

- anonymous

cos4t-sin4t=
kirb can u give me the whole solution soo i can studied it... please

- 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

Now use the double-angle formula

- kirbykirby

\[Since : \sin(2t)=2\sin t \cos t\]

- kirbykirby

\[-4\cos t \sin t = -2(2\cos t \sin t) = -2\sin(2t)\]

- anonymous

THANK YOU KIRB!!!

- 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

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

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

I'm usually the brute force guy. Once in a while I see one!

- kirbykirby

Hehe good one ;)

- anonymous

how about differentiate: y=sec²x-tan²x

- 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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