# Prove trig identity 2cos^2x-1=cos^4x-sin^4x [Solved!]

**Alexandra** 25 Nov 2015, 09:54

### My question

Comments: Please could you help with this problem?

2cosĀ²x-1=cos^4x-sin^4x

### Relevant page

1. Trigonometric Identities

### What I've done so far

Using the hints you gave on the page, I got this far:

RHS = cos^4x-sin^4x

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

But I'm stuck there

X

Comments: Please could you help with this problem?
2cos²x-1=cos^4x-sin^4x

Relevant page
<a href="/analytic-trigonometry/1-trigonometric-identities.php">1. Trigonometric Identities</a>
What I've done so far
Using the hints you gave on the page, I got this far:
RHS = cos^4x-sin^4x
= (cos^2x)^2-(sin^2x)^2
But I'm stuck there

## Re: Prove trig identity 2cos^2x-1=cos^4x-sin^4x

**Newton** 26 Nov 2015, 03:42

Hello Alexandra

Please use the math input system. It makes it a lot easier for us (and you) to read your math.

I am using the principles you found (from the page 1. Trigonometric Identities)

# Work on the most complex side and simplify it so that it has the same form as the simplest side.

# Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only.

# In most examples where you see power 2, it will involve using the identity `sin^2 theta + cos^2 theta = 1`.

Here's your problem:

`2cos^2 x-1=cos^4x-sin^4x`

Your rightly noticed the most complex side is the right hand, since we don't have `(cosx)^4` or `(sinx)^4` in our formulae.

What you wrote is a difference of 2 squares, from algebra:

2. Common Factor and Difference of Squares

Do you think you can go on from there using that hint?

Regards

X

Hello Alexandra
Please use the math input system. It makes it a lot easier for us (and you) to read your math.
I am using the principles you found (from the page <a href="/analytic-trigonometry/1-trigonometric-identities.php">1. Trigonometric Identities</a>)
# Work on the most complex side and simplify it so that it has the same form as the simplest side.
# Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only.
# In most examples where you see power 2, it will involve using the identity `sin^2 theta + cos^2 theta = 1`.
Here's your problem:
`2cos^2 x-1=cos^4x-sin^4x`
Your rightly noticed the most complex side is the right hand, since we don't have `(cosx)^4` or `(sinx)^4` in our formulae.
What you wrote is a difference of 2 squares, from algebra:
<a href="/factoring-fractions/2-common-factor-difference-squares.php">2. Common Factor and Difference of Squares</a>
Do you think you can go on from there using that hint?
Regards

## Re: Prove trig identity 2cos^2x-1=cos^4x-sin^4x

**Alexandra** 26 Nov 2015, 23:56

Hi,

I'll try to do the math input like you said.

`RHS = cos^4x-sin^4x`

`= (cos^2x)^2-(sin^2x)^2`

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

`= (cos^2 x - sin ^2 x)` because `sin^2 x + cos^2 x = 1`.

Then

`sin ^2 x = 1 - cos^2 x`

so

`RHS = (cos^2 x - (1 - cos^2 x))`

`= 2 cos^2 x - 1`

`= LHS`

Yay!

Can I ask some more?

X

Hi,
I'll try to do the math input like you said.
`RHS = cos^4x-sin^4x`
`= (cos^2x)^2-(sin^2x)^2`
`= (cos^2 x - sin ^2 x)(cos^2 x + sin ^2 x)
`= (cos^2 x - sin ^2 x)` because `sin^2 x + cos^2 x = 1`.
Then
`sin ^2 x = 1 - cos^2 x`
so
`RHS = (cos^2 x - (1 - cos^2 x))`
`= 2 cos^2 x - 1`
`= LHS`
Yay!
Can I ask some more?

## Re: Prove trig identity 2cos^2x-1=cos^4x-sin^4x

**Murray** 27 Nov 2015, 20:57

Good work!

Sure, you can ask more. Always have a go at it first and show us what you have done.

X

Good work!
Sure, you can ask more. Always have a go at it first and show us what you have done.

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