Question

use the square identities to simplify 2sin^2xsin^2x
I tried using sin^2 x =1-cos2x/2
which made the equation into
2(1-cos2x/2)(1-cos2x/2)
I am pretty sure I did this wrong. Someone please help and give me some pointers!!!

Answer 1

\[\large\rm 2\sin^2x \cdot\sin^2x\]This is what they want us to simplify? 0_o

Answer 2

What do they mean by "simplify"?
Do you want only first degree sines and cosines?
The directions are very unclear.

Answer 3

Well the options for the solutions are
3-4cos(2x)-cos(4x)/4
3+4cos(2x)-cos(4x)/4
3+4cos(2x)+cos(4x)/4
3-4cos(2x)+cos(4x)/4

Answer 4

I am just not seeing where any of the numbers are coming from considering we are using the initial equation 2sin^2xsin^2x

Answer 5

Yah I guess that's what they want then :)
Ok your first step looks great!
Making use of your Half-Angle Formula:\[\large\rm 2\sin^2x \cdot\sin^2x\quad=2\cdot\frac{1}{2}[1-\cos(2x)]\cdot\frac{1}{2}[1-\cos(2x)]\]

Answer 6

Let's bring all the fraction business to the front,\[\large\rm =\frac{1}{2}[1-\cos(2x)][1-\cos(2x)]\]We'll have to FOIL this out.

Answer 7

If we put the cosine stuff in the numerator here it might make things easier for us.\[\large\rm \frac{(1-\cos2x)^2}{2}=\frac{1-2\cos(2x)+\cos^2(2x)}{2}\]Get lost in those steps somewhere? :o
What do you lady?

Answer 8

What do you think? :)
typo lol

Answer 9

I think I see what has been done. but why include the half angle formula along with the squared identity?

Answer 10

hint use
\[\cos^2(u)=\frac{1+\cos(2u)}{2}\]
again

Answer 11

ohh so we are using the cos2x identity, not sin

Answer 12

cos^2x

Answer 13

well you see what @zepdrix has typed... things are in terms cos...
you want to write cos^2(2x) in terms of cos(4x) though
and that identity I mention will allow you do just that
replace u with 2x to see things better

Answer 14

okay I am getting it now

Answer 15

Ya, the first time you had to use the sine^2 identity,
then you foil out all the mess,
then you apply cosine^2 identity.

It's a little tricky ^^

Answer 16

okay that is going to be hard to remeber for other equations but practice makes perfect I guess

Answer 17

how do I get from here to the solution though?

Answer 18

Umm lemme write out the steps in a little cleaner way maybe.

Answer 19

We have to use cos^2x to solve right? but I dont get how that is plugged in

Answer 20

\[\large\rm 2\color{orangered}{\sin^2x} \cdot\color{orangered}{\sin^2x}\quad=2\color{orangered}{\left(\frac{1-\cos2x}{2}\right)} \cdot\color{orangered}{\left(\frac{1-\cos2x}{2}\right)}\]Cancel out a pair of 2's,
Expand out the numerator,\[\large\rm =\frac{1-2\cos2x+\color{royalblue}{\cos^22x}}{2}\]Applying our Cosine Half-Angle Formula gives us,\[\large\rm =\frac{1-2\cos2x+\color{royalblue}{\left(\dfrac{1+\cos4x}{2}\right)}}{2}\]

Oh yes, that step is a little tricky :o
I should back up maybe..

Answer 21

\[\large\rm \cos^2(\color{red}{x})\quad=\frac{1+\cos(2\color{red}{x})}{2}\]What ever angle you start with (in red),
you should end up with double that after you apply the formula.
So if instead we started with this,\[\large\rm \cos^2(\color{red}{2x})\quad=\frac{1+\cos(2\color{red}{\cdot2x})}{2}\]Do you see why we end up with 4x as the new angle?
We're doubling the 2x

Answer 22

got it

Answer 23

since its called the "double" angle identity

Answer 24

It's called the Half-Angle Formula :)
The angle is cut in half if you read from the other direction.

Some teachers reverse the half-angle and double angle formulas...
I find that confusing :(

Answer 25

okay now what?

Answer 26

@zepdrix

Answer 27

\[\large\rm =\frac{1-2\cos2x+\color{royalblue}{\cos^22x}}{2}\]So you apply your Half-Angle Formula to that cos^2(2x),\[\large\rm =\frac{1-2\cos2x+\color{royalblue}{\left(\dfrac{1+\cos4x}{2}\right)}}{2}\]From here, you `could` get a common denominator and all that fancy business.
But I would recommond, instead, that you multiply top and bottom by 2.\[\large\rm =\frac{1-2\cos2x+\left(\dfrac{1+\cos4x}{2}\right)}{2}\cdot\color{orangered}{\frac{2}{2}}\]Which gives us,\[\large\rm =\frac{2-4\cos2x+\left(1+\cos4x\right)}{2}\]And then dro the brackets, nothing is being applied to them.
\[\large\rm =\frac{2-4\cos2x+1+\cos4x}{2}\]And combine like-terms.
Which is only the 2 and 1 in this case.

Answer 28

Woops! 4 on the very bottom from multiplying by 2, my bad.\[\large\rm =\frac{2-4\cos2x+1+\cos4x}{4}\]

Answer 29

okay and then it would be
3-4cos2x_cos4x/4