Question

Verify/establish the identity:
[(2sin^2 theta-1)^2/(sin^4 theta-cos^4 theta)=1-cos^2 theta
I feel like I should work with the LHS in verifying the RHS.

Answer 1

\[\Large\rm \frac{(2\sin^2\theta-1)^2}{(\sin^4\theta-\cos^4\theta)}\]Ooo this is an interesting problem :)

Answer 2

Bottom we have the `difference of squares`, it might not be super easy to notice at first though.

Answer 3

Remember how to factor difference of squares into conjugates?
Looks like this:
\[\Large\rm a^4-b^4=(a^2)^2-(b^2)^2=(a^2-b^2)(a^2+b^2)\]

Answer 4

After that, I would probably apply the `Cosine Double Angle Formula` to top and bottom.

Need to see a few more steps? :o

Answer 5

I might be wrong about the Double Angle, I'm trying to do too many steps in my head D:

Answer 6

If that's okay. I just learned about Double angle and half-angle formulas today and I'm still fairly uncertain. Thank you for helping me again!

Answer 7

Ok I applied 2 steps here,
\[\Large\rm =\frac{(1-2\sin^2\theta)^2}{(\sin^2\theta-\cos^2\theta)(\color{royalblue}{\sin^2\theta+\cos^2\theta})}\]In the numerator I factored a -1 out of each term (it get's squared when coming out of the brackets and becomes positive).
In the denominator we applied the difference of squares formula.
Notice anything interesting about the blue portion?

Answer 8

It seems like this is going to end up simplifying to \(\Large\rm 1-2\cos^2\theta\)
Are you sure you wrote it down correctly? :o
Again I'm doing these steps in my head >.< so I could be wrong...

Answer 9

Yes, the change in sign. Is that due to the difference of squares or is that the cosine double angle formula?

Answer 10

Hmm?
Are you talking about the numerator or denominator step?

Answer 11

Oh I asked about the blue part :) lol.
Yes that's due to the difference of squares formula.

Answer 12

Oh sorry, I forgot to type the two. I am so terribly sorry!

Answer 13

The denominator.

Answer 14

lolol you apologize too much ^^ simmer down ms Alyssa!! so proper! lol

Answer 15

Okay, just making sure.

Answer 16

Then how do you implement the cosine double angle formula from here, if I may ask.

Answer 17

That's not what I wanted you to notice about it though!! :)
The blue part is our Pythagorean Identity for sine and cosine yes?
That simplifies down REALLY nice.

Answer 18

Ohhhh I hadn't noticed that! I'm sorry, I'm really focused on the angle formulas.

Answer 19

I guess you don't have to apply the Cosine Double Angle Formula if you're uncomfortable with it.

It's just that we want to try and make these orange parts the same,\[\Large\rm =\frac{(\color{orangered}{1-2\sin^2\theta})^2}{(\color{orangered}{\sin^2\theta-\cos^2\theta})(\color{royalblue}{\sin^2\theta+\cos^2\theta})}\]

Answer 20

We can do that by simply using the Pythagorean Identity and changing our cos^2(theta) to 1-sin^2theta

Answer 21

Thank you! I'm sorry that I didn't see it before!

Answer 22

LOL there you go again XD

Answer 23

\[\Large\rm =\frac{(\color{orangered}{1-2\sin^2\theta})^2}{(\color{orangered}{\sin^2\theta-(1-\sin^2\theta)})(\color{royalblue}{1})}\]

Answer 24

Which leads us to,\[\Large\rm =\frac{(\color{orangered}{1-2\sin^2\theta})^2}{(\color{orangered}{2\sin^2\theta-1})}\]AHHH I shoulda left the top alone....
Remember when I factored out that negative earlier?
blah!

Answer 25

So put the negative back in XD\[\Large\rm =\frac{(\color{orangered}{2\sin^2\theta-1})^2}{(\color{orangered}{2\sin^2\theta-1})}\]

Answer 26

Yes, I remember. Don't worry, I'm just grateful that you're guiding me through this problem.

Answer 27

So if we haven't made any mistakes up to this point, it looks like we get a nice cancellation,\[\Large\rm =\frac{(2\sin^2\theta-1)^{\cancel2}}{\cancel{(2\sin^2\theta-1)}}\]

Answer 28

But as I mentioned before, this is not going to simplify to \(\Large\rm 1-\cos^2\theta\).
So I hope the problem was written down incorrectly D:

Answer 29

Yes, I'm sorry again. It's supposed to simplify to 1-2cos^2 theta

Answer 30

Thank you so much for being so patient with me.

Answer 31

Ah ok great :) that looks better.

Answer 32

So you're left with,\[\Large\rm 2\color{#CC0033}{\sin^2\theta}-1\]And you would just apply your Pythagorean Identity on this sin^2(theta) a final time to get your answer.

Any of those steps too confusing? :o

Answer 33

Like this,
\[\Large\rm =2(\color{#CC0033}{1-\cos^2\theta})-1\](Just in case there was any confusion with that step >.< )

Answer 34

Not anymore. Once I saw where you redistributed your factor. I really cannot thank you enough for all your help.

Answer 35

This problem is a bit of a doozy!
Lot of little things to remember, most importantly: Difference of Squares.

Answer 36

Aw great c: Glad to help!

Answer 37

Yeah, sometimes it's difficult to recognize where I need to apply difference of squares, but it comes with practice, right. :)
Thank you again! For being such a kind and thoughtful instructor!

Answer 38

G'day M'lady ^^/
See you at the next problem perhaps. hehe

Answer 39

I hope you have a wonderful day as well!
I would be overjoyed to have you guide me again!