Question

How did they factor this part?

Answer 1

Notice that due to product rule the sin^(n) x is still there and the next step its not.

Answer 2

why not?

Answer 3

for the second term: sin^n = sin ^(n-1+1) = sin^n-1 *sin, then factor, what's wrong?

Answer 4

I don't understand the part

Answer 5

ok, let wait for Psymon 's explanation. he is good at that. if you still get stuck, then ask him. to me, nothing wrong with that

Answer 6

just show me that expanded part and explain the exponent rule that occurred there.

Answer 7

When you factor out a negative exponent, you actually add that same negative power onto what you just factored it out of. For example, if I have

sin^-2(x)cosx + sin^3(x)tanx. If I factor out a sin^-2, I'll be adding a sin^2 to each term, which will make it this:

sin^-2(x)(cosx + sin^5(x)tanx). thats what happened with yours. You have an nsin^(n-1), which is the same as nsin^n(x) * sin^-1(x)So I'll rewrite and then show you the factoring.

\[nsin^{n}x*\sin^{-1}xcos(x)\cos(nx)- nsin^{n}xsin(nx)\]I just rewrote what your problem has but with the sin^(n-1) split and the -n moved to the front of the second term. Now from this view, clearly nsin^n(x) can be factored out, which would leave:

\[nsin^{n}x(\sin^{-1}xcos(x)\cos(nx) - \sin(nx))\]But apparently they didn't like that, so they also factored out the sin^-1 portion. Now as I just said, factoring out a negative power ADDS that same power of whatever back into what it is being factored out from. So if I go further and factor out the sin^-1, a sin^1(x) will be added into each term, which is where your answer comes in:

\[nsin^{n-1}x(\sin^{-1+1}xcos(x)\cos(nx)-\sin^{0+1}xsin(nx))\] Which becomes
\[nsin^{n-1}x(\cos(nx)\cos(x) - \sin(nx)sinx))\]

Answer 8

@mebs, now is my turn to fire you, hehehe...

Answer 9

That makes a lot of sense... It wouldn't be the same otherwise .

Answer 10

I already ate you bro. ;) @Loser66

Answer 11

Hope that helps then ^_^

Answer 12

@mebs so, to you, me died? ok,

Answer 13

No, your just another penguin. Your cousin is in my stomach. your he is being digested.

Answer 14

Alright thanks everybody.

Answer 15

O wait when you factored out that sin^n(x) that other part had no sin terms. So basically you had to put it there because it wouldn't have been the same otherwise @Psymon

Answer 16

Right. You kinda have to think of it as being a 0 power, meaning just equal to 1. Can do that with anything.
x + y. If I randomly factor out a x^-1, it becomes (x^2 + xy). You assume theres some sort of x^0 power there and you add 1 in.

Answer 17

basically to get rid of \[\sin^{-1}x \to * \frac{sinx}{sinx}\]

Answer 18

Well, I should correct what I put above and say x^-1(x^2 + xy)

Answer 19

wait no not that.. more like you just did \[\frac{1}{sinx} * sinx and than \to 1*sinx\]

Answer 20

so when you put it back it goes to being \[\sin^{-1}x \to 1 \]

Answer 21

Yeah, if I were to put it back, exactly, itd just revert back to being 1.

Answer 22

I GET IT! thanks

Answer 23

Awesome :3