Proving Trig Identities: Solving sin^4x + cos^4x = 1-2sin^2xcos^2x

[snooooooooooz]

I'm solving a pretty descent trig identity question, but I'm stuck. I'm not going to type out the original question, but here the section that I'm stuck on: sin^4x + cos^4x and here is what I have to prove: 1-2sin^2xcos^2x

I know that I'm really close, I just can't get this section. Any help is much appreciated.

 

[rocomath]

Show your work and you'll get a lot of help.

 

[rocomath]

To be honest, there is just too much to type out.

To be honest, I don't care. Wait for someone else :)

 

[snooooooooooz]

To be honest, I don't care. Wait for someone else :)

... I would like your help, I just don't know what to type out. Basically, I rearranged the equation using sine and cosine and I ended up with sin^4x + cos^4x/sinxcosx.

 

[rocomath]

What are you trying to do? Make the Left equal the Right?

 

[snooooooooooz]

What are you trying to do? Make the Left equal the Right?

Yes, you're proving that one side equals the other side.

 

[snooooooooooz]

From the first line my next line is:
sin^3x/cos^3x / 1/cos^2x + cos^3x/sin^3x / 1/sin^2x

then, sin^3x/cos^3x x cos^3x / 1/cos^2x x cos^3x + cos^3x/sin^3x x sin^3x / 1/sin^2x x sin^3x

then, sin^4x + cos^4x / sinx x cosx

so, as you can see, I'm stuck. You can't simplify the sum of two even powers, so I have clue what to do next.

 

[Gib Z]

We can simplify $\sin^4 x + \cos^4 x$ quite well actually, stick with what you're trying =]

We CAN simplify any polynomial expression with real coefficients into linear and quadratic terms, or if you don't mind complex coefficients, all linear terms. This is most easily been, and actually directly stated, by the Fundamental Theorem of Algebra, but also can be cleverly seen with a nice application of the conjugate root theorem =] It doesn't matter if you don't understand most of this by the way.

Perhaps you should try completing the square =] ?

 

[kamerling]

the only other thing you need is $$sin^2x + cos^2x = 1$$

 

[tiny-tim]

sin^4x + cos^4x and here is what I have to prove: 1-2sin^2xcos^2x

Hi snooooooooooz! :zzz:

Hint: you know what sin^2x + cos^2x is, don't you?

Well, how can you use that to help with sin^4x + cos^4x?

 

[snooooooooooz]

EDIT: Actually, I don't see how you can use completing the square to factor sin^4x + cos^4x.

 

[tiny-tim]

EDIT: Actually, I don't see how you can use completing the square to factor sin^4x + cos^4x.

snooooooooooz, stop trying to factorise it!

The object is to simplify it.

Use the hint that kamerling and I gave you:

sin^2x + cos^2x = 1​

 

[snooooooooooz]

snooooooooooz, stop trying to factorise it!

The object is to simplify it.

Use the hint that kamerling and I gave you:

sin^2x + cos^2x = 1​

ok so if I take sin^2x + cos^2x out of sin^4+cos^4x that means I'm left with sin^2x + cos^2x. Since sin^2x + cos^2x = 1 and cos^2x = 1-sin^2x there is NO cos^2x at the end to multiply by. There is no way I can see how sin^4x + cos^4x can equal 1-2sin^2x cos^2x

 

[kamerling]

try to write both expressions using just $$sin^2x$$

 

[tiny-tim]

… BIG hint …

Hint: what is (sin^2x + cos^2x)^2 ?

 

[snooooooooooz]

Hint: what is (sin^2x + cos^2x)^2 ?

ok.. so my next line is: $$1$$ x $$sin^2x +cos^2x$$?

 

[tiny-tim]

No!

Your next line is: (sin^2x + cos^2x) x (sin^2x + cos^2x)

 

[snooooooooooz]

No!

Your next line is: (sin^2x + cos^2x) x (sin^2x + cos^2x)

But if you foil that out you get sin^4x + sin^2xcos^2x + cos^2xsin^2x + cos^4x.

 

[snooooooooooz]

So I got $$sin^4x +2sin^2xcos^2x + cos^4x$$. How would drop the $$sin^4x$$ and $$cos^4x$$ to make that into a 1-?

 

[tiny-tim]

But if you foil that out you get sin^4x + sin^2xcos^2x + cos^2xsin^2x + cos^4x.

(what's "foil"?)

Yes! Or, slightly simpler: sin^4x + cos^4x + 2sin^2xcos^2x.

Now your next line is:
So sin^4x + cos^4x + 2sin^2xcos^2x = … ?

 

[snooooooooooz]

(what's "foil"?)

Yes! Or, slightly simpler: sin^4x + cos^4x + 2sin^2xcos^2x.

Now your next line is:
So sin^4x + cos^4x + 2sin^2xcos^2x = … ?

Sorry, "foil" is distributive property. I'm just going to assume that sin^4x + cos^4x = 1- but how?

 

[tiny-tim]

I'm just going to assume that sin^4x + cos^4x = 1- but how?

Because 1 = (sin^2x + cos^2x)^2 = sin^4x + cos^4x + 2sin^2xcos^2x;

so sin^4x + cos^4x = 1 - 2sin^2xcos^2x.

So your next line is … ?

 

[snooooooooooz]

Because 1 = (sin^2x + cos^2x)^2 = sin^4x + cos^4x + 2sin^2xcos^2x;

so sin^4x + cos^4x = 1 - 2sin^2xcos^2x.

So your next line is … ?

but where did the negative come from?

 

[tiny-tim]

but where did the negative come from?

ah! algebra difficulties!

hmm … if 1 = A + B + C,

then A + B = 1 - C.
​

Yes? No? You have to be able to do these!

 

[Jamil (2nd)]

If you just let: sin²x = a
and: cos²x = b

then,
a² = sin⁴
and
b² = cos⁴

thus,
a² + b² + 2ab = (a + b)²

or, a² + b² + 2ab - 2ab = (a + b)² - 2ab

or, a² + b² = (a + b)² - 2ab

then what?