Request for help proving this identity

[desertwonder]

I've wracked my brains out trying to prove this identity. If anyone could offer some suggestions, I'd greatly appreciate it!

(sin(x))^4+(cos(x))^4 = 0.25∙cos(4x)+0.75

 

[HallsofIvy]

cos(2x)= cos²(x)- sin²(x)
sin(2x)= 2sin(x)cos(x)

So cos(4x)= ?

 

[desertwonder]

Cos(4x)=Cos(2(2x))=
...So does that mean Cos(4x)=2((cosx)^2-(sinx)^2)?

 

[HallsofIvy]

Cos(4x)=Cos(2(2x))=
...So does that mean Cos(4x)=2((cosx)^2-(sinx)^2)?

Only if x² is the same as 2x.

 

[desertwonder]

Hmmm...in that case...I'm not really sure how to go about reducing cos(4x) to a single angle using the double-angle cosine formula.

EDIT: Ahh...I now see that cos(4x)=cos(2x+2x)=cos(2x)cos(2x)-sin(2x)sin(2x)...
more work underway!

EDIT2: Ohhh, here we go: cos(4x)=((cos(x))^2- (sin(x))^2)((cos(x))^2- (sin(x))^2)-(2sin(x)cos(x))(2sin(x)cos(x))!

 

[desertwonder]

Hmm...this was all I could narrow it down to:

cos(4x)= (cos(x))^4 + (sin(x))^4 - 6(sin(x))^2(cos(x))^2

 

[HallsofIvy]

Hmm...this was all I could narrow it down to:
cos(4x)= (cos(x))^4 + (sin(x))^4 - 6(sin(x))^2(cos(x))^2

Good! That's exactly right.

Now go back to the original identity. Because I dislike fractions, multiply through by 4. Then put formula you have above into the equation and move (almost) everything over to the left side.

 

[desertwonder]

Hmm...after fiddling around some more following your suggestions I have:

(sin(x))^4+(cos(x))^4 = = -2(sin(x))^2(cos(x))^2

 

[HallsofIvy]

Hmm...after fiddling around some more following your suggestions I have:
(sin(x))^4+(cos(x))^4 = = -2(sin(x))^2(cos(x))^2

What happened to the "3/4" in the orginal equation??

 

[desertwonder]

The 3/4 turned into a 3 after I multiplied everything by 4. Afterwards, the 4sin and 4cos on the left became 3sin and 3cose when moving the stuff on the right to the left. I then divided everything by 3.

 

[VietDao29]

When divided by 3, shouldn't it become '1'?
$$3 \cos ^ 4 x + 3 \sin ^ 4 x = -6 \sin ^ 2 x \cos ^ 2 x + 3$$
Devide both sides by 3 gives:
$$\Leftrightarrow \cos ^ 4 x + \sin ^ 4 x = -2 \sin ^ 2 x \cos ^ 2 x + 1$$

 

[HallsofIvy]

The 3/4 turned into a 3 after I multiplied everything by 4. Afterwards, the 4sin and 4cos on the left became 3sin and 3cose when moving the stuff on the right to the left. I then divided everything by 3.

Yes! Now where is the 1 you get when you divide by 3?

Look at VietDao's post.
After moving (almost) everything to the left you should have:
$$cos^4(x)+ 2sin^2(x)cos^2(x)+ sin^4(x)= 1$$

Does that give you any ideas? How can you factor the left hand side?