Question

Simplify using the squared identities.

Answer 1

\[2\sin^2x \cos^2x\]

Answer 2

Double-angle formula:
sin2x = 2 * sinx * cosx

Answer 3

Hm... I think I have trouble going on from there.

Answer 4

heres the original

Answer 5

They consider those options simplified? lol
Ok anyway, let's try this maybe.\[\Large 2(\sin x \cos x)^2\]From here we'll apply `Sine Double Angle`,\[\Large 2\left(\frac{1}{2}\sin(2x)\right)^2\]Multiplying out our square gives us,\[\Large \frac{1}{2}\sin^2(2x)\]Our `Square Identity` gives us,\[\Large \frac{1}{2}\left(1-\cos^2(2x)\right)\]Hmmmmmmm

Answer 6

thats double angle?

Answer 7

Oh I guess we could apply the `Cosine Half Angle Formula` from that point.\[\large \color{royalblue}{\cos^2(x)=\frac{1}{2}(1+\cos(2x))}\]

So if we have cos^2(2x), it will give us,\[\large \color{royalblue}{\cos^2(2x)=\frac{1}{2}(1+\cos(4x))}\]

Applying this identity to our problem,\[\Large \frac{1}{2}\left(1-\cos^2(2x)\right) \qquad=\qquad \frac{1}{2}\left(1-\frac{1}{2}(1+\cos(4x))\right)\]

Answer 8

Wow this problem is a pain in the butt lol.

Answer 9

The Sine Double Angle Step? Was that a little confusing the way I did that?

Answer 10

well thatdouble angle formula looks different from what im used to

Answer 11

\[\Large 2\sin x \cos x=\sin(2x)\]But we're missing a 2, so we end up dividing by 2,\[\Large \sin x \cos x=\frac{1}{2}\sin(2x)\]

Answer 12

ohhh i didnt know you can do that

Answer 13

I find it funny how they say "Use square identities"...
Maybe I'm missing something but it seems like we have to use a ton of identities for this problem XD Not just square identity lol

Answer 14

Yeah, and it seems so similar to that double-angle formula. . . mah gash. . . ~_~

Answer 15

oh wait the originial problem has sin squared and cosine squared;
THIS is the problem:

\[2 \sin^2x \cos^2x \]

Answer 16

Oh oh it actually makes a lot more sense to apply the `Half-Angle Formula` from this point,\[\Large \frac{1}{2}\sin^2(2x)\]

Answer 17

so you take out the squares?

Answer 18

Yes I rewrote it,\[\Large 2\sin^2x \cos^2x=2(\sin x \cos x)^2\]

Answer 19

Sort of like this,\[\Large 2x^2y^2=2(xy)^2\]

Answer 20

oh i see

Answer 21

`Sine Half-Angle Formula`:\[\large \color{royalblue}{\sin^2x=\frac{1}{2}\left(1-\cos(2x)\right)}\]

So we can use this in our problem from this point,\[\Large \frac{1}{2}\sin^2(2x) \qquad=\qquad \frac{1}{2}\left[\frac{1}{2}\left(1-\cos(2x)\right)\right]\]

Answer 22

Woops*
\[\Large \frac{1}{2}\sin^2(2x) \qquad=\qquad \frac{1}{2}\left[\frac{1}{2}\left(1-\cos(4x)\right)\right]\]

Answer 23

Super confused by any of this?
We've had to apply like 3 different identities to get it to this point :( easy to get lost

Answer 24

Hm. . . so looking at it this way, the answer would be B, correct?

Answer 25

sorta LOL

Answer 26

i kinda lost you after \[2(sinx \cos x)^2\]

Answer 27

oh XD lol

Answer 28

i believe you use double angle?

Answer 29

it makes more sense to me

Answer 30

Take your identity,\[\large \color{royalblue}{2\sin x \cos x=\sin(2x)}\]Divide both sides by 2,\[\large \color{royalblue}{\sin x \cos x=\frac{1}{2}\sin(2x)}\]

So how can we apply this identity in it's currect form to our problem? :O\[\Large 2\left(\color{royalblue}{\sin x \cos x}\right)^2\]

Answer 31

\[2(1/2 \sin2x)^2\]

Answer 32

Good good :3
So from there, we need to square everything that's inside the brackets.
(1/2)^2 and also square the sine function.

Answer 33

does it become \[2(1/4\sin^22x^2)?\]

Answer 34

Woops, :O Don't square the inside of the sine function. :D

Answer 35

\[\Large 2\cdot\frac{1}{4}\sin^2(2x)\]Leave that 2x alone! :)

Answer 36

oh LOL