SOLUTION: Hello, The problem that I have is that I need to verify the identity: sin^4x+cos^4x=1-2cos^2x+2cos^4x I started with the right hand side and I tried to group the two co

Algebra ->  Trigonometry-basics -> SOLUTION: Hello, The problem that I have is that I need to verify the identity: sin^4x+cos^4x=1-2cos^2x+2cos^4x I started with the right hand side and I tried to group the two co      Log On


   

Question 635024: Hello,
The problem that I have is that I need to verify the identity:
sin^4x+cos^4x=1-2cos^2x+2cos^4x
I started with the right hand side and I tried to group the two cosine identities
=1-2cos^2x+2cos^4x
=1-2cos^2x(1-cos^2x)
using the Pythagorean identity I got
=1-2cos^2x(sin^2x)
I'm not sure if I can use the Pythagorean identity on the first part making it
=(2sin^2x)(sin^2x)
But if I do, I'm not getting any closer to the left hand side.
Please help, or tell me if I'm at least on the right track.
Thank you
Found 2 solutions by richwmiller, MathTherapy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
You have cos^4x on both sides.
so subtracting it from both sides
sin^4x=1-2cos^2x+cos^4x
apparently this is a known identity.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
The problem that I have is that I need to verify the identity:
sin^4x+cos^4x=1-2cos^2x+2cos^4x
I started with the right hand side and I tried to group the two cosine identities
=1-2cos^2x+2cos^4x
=1-2cos^2x(1-cos^2x)
using the Pythagorean identity I got
=1-2cos^2x(sin^2x)
I'm not sure if I can use the Pythagorean identity on the first part making it
=(2sin^2x)(sin^2x)
But if I do, I'm not getting any closer to the left hand side.
Please help, or tell me if I'm at least on the right track.
Thank you

sin%5E4%28x%29+%2B+cos%5E4%28x%29+=+1+-+2cos%5E2%28x%29+%2B+2cos%5E4%28x%29

sin%5E4%28x%29+=+1+-+2cos%5E2%28x%29+%2B+2cos%5E4%28x%29+-+cos%5E4%28x%29

sin%5E4%28x%29+=+1+-+2cos%5E2%28x%29+%2B+cos%5E4%28x%29

Since sin%5E2%28x%29+=+1+-+cos%5E2%28x%29, then sin%5E4%28x%29+=+%281+-+cos%5E2%28x%29%29%5E2

We now have: highlight_green%28%281+-+cos%5E2%28x%29%29%5E2+=+1+-+2cos%5E2%28x%29+%2B+cos%5E4%28x%29%29

When the left-side is FOILED, you end up with the right-side.

OR

sin%5E4%28x%29+%2B+cos%5E4%28x%29+=+1+-+2cos%5E2%28x%29+%2B+2cos%5E4%28x%29

sin%5E4%28x%29+=+1+-+2cos%5E2%28x%29+%2B+2cos%5E4%28x%29+-+cos%5E4%28x%29

sin%5E4%28x%29+=+1+-+2cos%5E2%28x%29+%2B+cos%5E4%28x%29

2cos%5E2%28x%29+-+cos%5E4%28x%29+=+1+-+sin%5E4%28x%29

cos%5E2%28x%29%282+-+cos%5E2%28x%29%29+=+1+-+sin%5E4%28x%29

cos%5E2%28x%29%282+-+cos%5E2%28x%29%29+=+%281+-+sin%5E2%28x%29%29%281+%2B+sin%5E2%28x%29%29

Since 1+-+sin%5E2%28x%29+=+cos%5E2%28x%29, then: cos%5E2%28x%29%282+-+cos%5E2%28x%29%29+=+%281+-+sin%5E2%28x%29%29%281+%2B+sin%5E2%28x%29%29 becomes: cos%5E2%28x%29%282+-+cos%5E2%28x%29%29+=+cos%5E2%28x%29%281+%2B+sin%5E2%28x%29%29

This means that 2+-+cos%5E2%28x%29+=+1+%2B+sin%5E2%28x%29, or sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+2+-+1

highlight_green%28sin%5E2%28x%29+%2B+cos%5E2%28x%29+=+1%29 (TRUE PYTHAGOREAN IDENTITY)

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