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

### My question

2cos²x-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

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.

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.

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

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!

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

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.