Question

Help proving the identity sin^2xcos^2x=(1-cos4x)/8? Medal and fan

Thanks(:

Answer 1

@mathslover could you help me with this one? Unfortunately, this was a huge practice packet and I'm stuck on this question as well.

Answer 2

Yes, I will try my best @seehearcreate .

Answer 3

Awesome, thankyou. I started off with multiplying the squared identities on the left, is that correct?

Answer 4

So, we have :

\[\sin^2 (x) \cos^2 (x) = \cfrac{(1 - \cos (4x))}{8} \]

We will first look at LHS.
In the LHS, \(\sin^2 (x) \) can be written as \(1 -\cos^2 (x)\) . Right?

Answer 5

Right. Would we also write the squared identity of cos^2x?

Answer 6

By the way, the squared identity of sin^2x is (1-cos^2x)/2, not (1-cos^2)...

Answer 7

Oh wait, nevermind, I was thinking of something else. Continue.

Answer 8

Now,

we have :

\((1- \cos^2 (x) )(\cos^2 (x)) = \cos^2 (x) - \cos^4 (x) \)

Right?

Answer 9

Yes(:

Answer 10

So, now,

in RHS you had : \(\cfrac{1 - \cos^4 ( x) }{ 8}\)

Answer 11

No, it's (1-cos(4x))/8

Answer 12

Oh.. Sorry!

You need to simplify cos(4x) then.

Answer 13

So (2cos^2(2x)-1)?

Answer 14

Good work.

Answer 15

But simplify it further too.

Answer 16

Okay so 2(2cos^4x-1)-1)?

Answer 17

I will get back to you .. have to leave for dinner. May be @ganeshie8 will like to continue from here.

Sorry

Answer 18

Oh no, that's fine, thankyou for your help

Answer 19

We have

... sin² x. cos² x

= (1/4) ( 4 sin² x. cos² x )

= (1/4) ( 2 sin x. cos x )²

= (1/4) ( sin 2x )²

= (1/4) sin² 2x

= (1/4) { [ 1 - cos ( 2• 2x ) ] / 2 }

= (1/8)( 1 - cos 4x

Answer 20

Okay, I really don't understand that since you didn't really break it down into distinct steps. If you can add on to what @mathslover was trying to show me, that would be most helpful because that's the work I have down on my sheet.

Answer 21

@Loser66 Could you help me with this?

Answer 22

@Harsha19111999 gives you the whole thing.
Now I just walk you through follow his step, ok?
I times both sides by 4 to get \(4 sin^2(x)cos^2(x) = \dfrac{1-cos(4x)}{2}\) OK?

Answer 23

Alright, but wouldn't it but you would have to have it set up like (1=cos(4x))/2 * (cos^2x) correct?

Answer 24

Be patient and confirm if you get it.

Answer 25

But you would*

Answer 26

I do understand that since you multiped the squared sine identity by 4.

Answer 27

Hey didn't you get the answer? Should i explain again?

Answer 28

And we have sin(2x) = 2 sin x cos x
so that \((sin (2x))^2 = 4 sin^2(x) cos ^2(x) \) this is our left hand side, right? so that the left hand side can be written as \(sin^2(2x)\) , right?

Answer 29

No, I really didn't understand it, I needed it broken down into individual steps. Math is my slower subject, unfortunately.

Answer 30

From which step do you want?

Answer 31

Yes, I understand that.

Answer 32

The whole thing really. That's why I asked @Loser66 since he breaks down each step until it draws to the conclusion.

Answer 33

\[\cos 2x=\cos \left( x+x \right)=\cos x~\cos x-\sin x~\sin x=\cos ^2x-\sin ^2 x\]
\[=1-\sin ^2x-\sin ^2x=1-2 \sin ^2x \]
\[1-\cos 2x=2\sin ^2x\]
\[1-\cos 4 x=2 \sin ^2~2x=2\left( 2 \sin x~\cos x \right)^2=8 \sin ^2x~\cos ^2x\]
\[\frac{ 1-\cos 4x }{ 8 }=\sin ^2~x \cos ^2~x \]

Answer 34

Once again, I don't want just the answer, I want the steps explained...I don't understand it otherwise.

Answer 35

Can we continue?? Can you get my stuff?

Answer 36

I said we could continue..

Answer 37

now, we have the left hand side is \(sin^2(2x) \) right?

Answer 38

Yes, correct.

Answer 39

I rewrite it:
\[\color{red}{sin^2(2x)}= \color {blue}{\dfrac{1-cos (4x)}{2}}\]
From the left hand side, I manipulate a little bit. But first off, I want you to know something
@joanip_ it's a proving problem, not solving

Answer 40

That's what @mathslover was doing earlier.

Answer 41

Right, I understood that ^

Answer 42

\[sin^2 (\color{red}{x}) = \dfrac{1-cos (\color{red}{2x})}{2}\] . This is identity. OK?

Answer 43

and pay attention: the left hand side is x and the right hand side is 2x. OK?

Answer 44

Yeah, I got that. I know the identities.

Answer 45

so, if the left hand side is 2x, then the right hand side is 4x, right?

Answer 46

Correct.

Answer 47

\[sin^2(2x)=\dfrac{1+cos(4x)}{2}\] , ok?

Answer 48

Yep, I understand that completely.

Answer 49

good. hihihihi We are done, right?

Answer 50

No? the right side was divided by 8?

Answer 51

When we multiplied the right side by 4, did the 8 reduce down to 2 then?

Answer 52

We manipulate a little bit from the beginning. Re read it

Answer 53

For example: if seehearcreate = Loser 66
and seehearcreate = @dan815 so that, instead of proving seehearcreate = Loser 66, I can prove dan815 = Loser 66 and make conclusion that we are all equal hahahaha

Answer 54

Yeah, I got you. I forgot about multiplying the right side by 4.

Answer 55

Thankyou again(:

Answer 56

ok,