Question

Calculate the derivatives \(sin^{'}(x) = \frac{d}{dx}sin(x)\) and \(cos^{'}(x) = \frac{d}{dx}cos(x)\)
by differentiating the summands of the power series. Show as a corollary the identity
\(sin^{2}(x)+cos^{2}(x) = 1\) for all \(x \in \mathbb{R}\).

Answer 1

i get such a help but i dont know how to apply it to question
"the derivative of each of the terms then simplify and see what the power series now resembles "

Answer 2

So the power series for \(\sin(x)\) is \[\large \sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]And the power series for \(\cos(x)\) is\[\large \sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\]

Answer 3

Now, look at \[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\]what is the derivative of this?

Answer 4

hmm what would you say what the derivate of this?

Answer 5

Try to use the power rule on that to find the derivative. That is, \[\frac{d}{dx}(x^n)=nx^{n-1}\]Once you've looked at that, try to do it in general. What is the derivative of \[\frac{(-1)^nx^{2n+1}}{(2n+1)!}\]

Answer 6

i am so bad i cant start with anthing George..

Answer 7

\[(-1)^{n} \] is not convergenze i know :)

Answer 8

It's not too bad. Since you're pressed for time, let's start with the derivative of \[\frac{(-1)^nx^{2n+1}}{(2n+1)!}\]Using the power rule (we can ignore the -1), we get that\[\frac{d}{dx}\left(\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)=\frac{(2n+1)(-1)^nx^{2n}}{(2n+1)!}\]You can cancel the \(2n+1\) from the top and bottom to get\[\frac{(-1)^nx^{2n}}{(2n)!}\]This is exactly the power sum for \(\cos(x)\), so this shows that \(\frac{d}{dx}(\sin(x))=\cos(x)\).

Answer 9

ok..

Answer 10

this end solution George ? or you want to add more things?

Answer 11

i think it was end solution thank you very much @KingGeorge

Answer 12

Now for the other part\[\frac{d}{dx}\left(\frac{(-1)^nx^{2n}}{(2n)!}\right)=\frac{2n(-1)^nx^{2n-1}}{(2n)!}\]Once again, we can cancel the \(2n\), and get \[\frac{(-1)^nx^{2n-1}}{(2n-1)!}.\]Now, notice what happens if we subsitute \(n\) for \(n+1\). We get \[-\frac{(-1)^nx^{2n+1}}{(2n+1)!}\]

Answer 13

aah ok..

Answer 14

The sum of this is exactly equal to \(-\sin(x)\), so we have proved that \(\frac{d}{dx}(\cos(x))=-\sin(x)\).

Answer 15

ok thank you very much George, you are my live saver again.. i have 3 or 4 more questions i will post above, they are also not so difficult ones like the other weeks, if you like you can look at them

Answer 16

I can take a look at them. Also, we still haven't done the corollary, but here's how to do it.

Take the derivative of \(\sin^2(x)+\cos^2(x)\). Show that this is equal to 0. After that, substitute \(x=0\), to get a value of 1. Since the derivative is 0, the function is constant, and since it equals 1 at \(x=0\), it must be that \(\sin^2(x)+\cos^2(x)=1\).

Answer 17

sorry i thought its finished, pls feel free to write more if we are still not finished with it

Answer 18

The derivative of \(\sin^2(x)+\cos^2(x)\) is \[2\sin(x)\cos(x)+2\cos(x)(-\sin(x))=2\sin(x)\cos(x)-2\sin(x)\cos(x)=0\]

Answer 19

ok..

Answer 20

Then, at \(x=0\) we have \[\sin^2(0)+\cos^2(1)=0+1^2=1\]Thus, we have the identity \[\sin^2(x)+\cos^2(x)=1\]Since it must be constant for all \(x\in\mathbb{R}\)

Answer 21

also sin value is always like zero

Answer 22

The sin value is not always 0. However, the value of \(\sin^2(x)+\cos^2(x)\) is always 1. Remember, we were differentiating the entire function, and not just \(\sin(x)\).

Answer 23

ok..

Answer 24

Also, something just came up, and I'll have to be gone for about 20 minutes. So for now, you might want to get someone else to help you since I won't be here for a bit.

Answer 25

and this is our end solution?

Answer 26

Yes, this is the end solution.

Answer 27

ok thank you very much George, i write them clear on paper now