Question

how to solve 1/1-cos2x?

Answer 1

solve what ? ! an equation ?!

Answer 2

how do you verify that 1/1-cos2x is \[\left(\begin{matrix}1 \\ 2\end{matrix}\right)\] csc\[^{2}\] x

Answer 3

1/2csc²x

Answer 4

kk
use this \[\cos(2x)=1-2\sin^2(x)\]
calculate first 1-2cos(2x)
what did you get ?

Answer 5

1/sin²x?

Answer 6

sorry ;) 1-cos(2x) = ?! (I' typed 1-2cos(2x))

Answer 7

i dont get it...

Answer 8

cos2x= 1-2sin²x

Answer 9

\[\text{using this } \color{green}{\cos(2x)}=1-2\sin^2(x) \text{ what is } 1-\color{green}{\cos(2x)}?\]

Answer 10

1-1-sin²x...?

Answer 11

\[1-(1-2\sin^2(x))\]

Answer 12

oh okay i got it

Answer 13

what did you get .? :)

Answer 14

it would be 1/ 1-(1-2sin²(x))

Answer 15

But i dont know what to do on the next step.....

Answer 16

C'moon !

Answer 17

so what's \[-(1-2\sin^2(x)) \text{ without using this "(" ")"}\]

Answer 18

-cos2x

Answer 19

okay i am really lost doesnt it become -2sin²x?

Answer 20

That's not what I'm asking for ! what's -(a-b) without using this "(" ")"

Answer 21

-a-b

Answer 22

no ! -a+b
An example -(4-1)=-4+1=-3

Answer 23

OHHH yes yes yes i got what you mean

Answer 24

now ! if you are convinced ! what's \[1-(1-2\sin^2(x))\]

Answer 25

1-1+2sin²x

Answer 26

which is 2sin²x !!!!???

Answer 27

yes ! then 1/1-cos(2x) =?

Answer 28

are you there @lsugano

Answer 29

um so do we use this identity sinx = 1/cscx

Answer 30

yes !

Answer 31

what did you get ?

Answer 32

i dont know how the answer is 1/2cscx

Answer 33

is this step correct? http://bamboodock.wacom.com/doodler/2aff60b5-e920-4961-bb39-e9ab34c722b6

Answer 34

C'mooon ! \[\text{we said that } 1-\cos(2x)=2\sin^2(x)\]
then \[\frac{1}{1-\cos(2x)}=\frac{1}{2\sin^2(x)}=\frac{1}{2}(\frac{1}{\sin(x)})^2\] do you see that .

Answer 35

Soryy :'( I didn't see your response !

Answer 36

Okayy
hahaha i am really sorry i made this very difficult. i am a bad math learner

Answer 37

BUT i got what this means!!!
THank you so much!

Answer 38

what you did is also right !

Answer 39

so it becomes 1/2csc²x!!!!

Answer 40

yes ! \[\frac{1}{2} \csc^2(x) \]

Answer 41

oohh thank you so much! i couldnt have gone through this without you, seriously thank you thank you!!!

Answer 42

I checked your profile :D ! WELCOME TO OPENSTUDY !
you're welcome anytime ! (Lisa) :) :)

Answer 43

Thank you ! yeah, im really new to this so I hope i get used to it!! thank you !!

Answer 44

Yw :) if you need Anything else just send me the link of your question Or mention me ( @Neemo) !

Answer 45

Okay thank you very much!

Answer 46

i have another question

Answer 47

kk ! :)

Answer 48

How do you solve sin⁴a=(sin²a)²

Answer 49

so far i know that its \[\left(\begin{matrix}1-\cos2a \ \div\ 2\end{matrix}\right)^{2}\]

Answer 50

kk you mean this \[\sin^4(x)=(\sin^2(x))^2\] ?

Answer 51

yes!!!

Answer 52

that's not obvious for you !
\[(a^2)^2=?\]

Answer 53

(2²)²=16 like that?

Answer 54

i dont know how variable work with those.....

Answer 55

maybe ! but ! in general case we have this property
\[(a^n)^p=a^{np}\]

Answer 56

okay!

Answer 57

your example ! \[(2^2)^2=2^{(2).(2)}=2^4=16\]
so yes !

Answer 58

so what do we do next for this one?

Answer 59

we square 1-cos2x/2 right

Answer 60

It obvious !
\[(\sin^2(x))^2=(\sin(x))^4=\sin^4(x)\]

Answer 61

why you want to do that ?!

Answer 62

the question says write sin⁴x in terms of the first power of one or more cosine functions.
using power reducing identites. and sin²x=1-cos2x/2

Answer 63

ohh ! you didn't say that :) :) ?!

Answer 64

so since sin⁴x=(sin²x)²=====(1-cos2x/2)²

Answer 65

im sorry I didn't mention that !!

Answer 66

No prblem at all ! yes !go on I'm watching you ! :)

Answer 67

so is it like 1-cos2x² / 4

Answer 68

remember ! that you 'll go to jail if you ever do that again !this is the property !!
\[(a-b)^2=a^2-2ab+b^2 \]
and remember \[a^2-b^2=(a-b)(a+b) \text{ \not what you typed } ==////===(a-b)^2\]

Answer 69

so correct your mistake :) @lsugano

Answer 70

i will write down on a tablet and could you check if im right?

Answer 71

I'll be happy :)

Answer 72

http://bamboodock.wacom.com/doodler/c36aa285-89d4-4cb9-9b6f-c067645b055b

Answer 73

were you able to see it?

Answer 74

yes ! But It's you shouldn't have \[\cos^2(2x)\]

Answer 75

sorry for being late :) !

Answer 76

@lsugano how can you deal with that ?!

Answer 77

yes!
okay i think i got the answer!!

Answer 78

because you reminded me about (a-b)squared, this equation was really easy to figure out

Answer 79

can I check it ! caus that isn't supposed to be the final answer ! :)

Answer 80

3-4cos2x+cos4x/8

Answer 81

Sorry ! I underestimated you ! you're really smart :)
That's correct :) !

Answer 82

yay!

Answer 83

thank you somuch! you a great teacher!!

Answer 84

No !usually, I'm a very bad teacher ! just you're the special one !
another question ?

Answer 85

okay so i got sin²xcos⁴x

Answer 86

yeah ! what is the question ?!

Answer 87

its the same thing

Answer 88

and my first step i did was this

Answer 89

kk ! go on !

Answer 90

and i got this one

Answer 91

correct ! go on !

Answer 92

are you thre ?! @lsugano

Answer 93

ssryy im am almost done!!

Answer 94

could you please check?