Question

DERIVATIVES OF TRIGO IDENTITIES


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

Answer 1

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?

Answer 2

differentiate Y=cos4 t -sin4 t

Answer 3

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

Answer 4

We're looking for \(\dfrac{dY}{dt}\)?

Can you find \(\dfrac{d}{dt}\cos^{4}(t)\)

Answer 5

yes

Answer 6

What do you get for that?

Answer 7

ist assingment dude

Answer 8

Conversation Overlap Misunderstanding...

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

Answer 9

its our assignment in differential calculus...

Answer 10

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

Answer 11

the second one

Answer 12

i dont know how to type that

Answer 13

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

Answer 14

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

Answer 15

the book has an answer of -2sin 2 t

Answer 16

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

Answer 17

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

Answer 18

so now: \[-4\cos^3t(\sin t) - 4\sin^3t(\cos t) = -4\cos t*\sin t(\cos^t+\sin^2t)=-4\cos t*\sin t(1) \]

Answer 19

Now use the double-angle formula

Answer 20

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

Answer 21

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

Answer 22

THANK YOU KIRB!!!

Answer 23

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.

Answer 24

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).

Answer 25

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

Answer 26

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

Answer 27

Hehe good one ;)

Answer 28

how about differentiate: y=sec²x-tan²x

Answer 29

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.