Question

Prove cos^4 X +sin^4 x= 2cos^2 x for any acute angle​

Answer 1

Answer:

Proof provided in the explanation.

Step-by-step explanation:

$\cos^4(\theta)-\sin^4(\theta)+1$

I'm going to rewrite that middle term so I can apply Pythagorean Identity, $\sin^2(\theta)=1-\cos^2(\theta)$:

$\cos^4(\theta)-(\sin^2(\theta))^2+1$

$\cos^4(\theta)-(1-\cos^2(\theta))^2+1$

Use $(a+b)^2=a^2+2ab+b^2$:

$\cos^4(\theta)-[1-2\cos^2(\theta)+\cos^4(\theta)]+1$

Distribute:

$\cos^4(\theta)-1+2\cos^2(\theta)-\cos^4(\theta)+1$

Combine like terms:

$\cos^4(\theta)-\cos^4(\theta)-1+1+2\cos^2(\theta)$

Applying inverse property of addition:

$0+0+2\cos^2(\theta)$

Applying identity property of addition:

$2\cos^2(\theta)$

Answer 2

To prove cos^4x-sin^4x +1= 2 cos^2 x

cos^4x -sin^4 x+1

= ( cos^2x +sin^2x)(cos^2x - sin^2x) +1

You need to remember 3 basic results

sin^2x + cos^2x= 1

cos2x = cos^2x - sin^2x

1+ cos 2x = 2 cos^2x

Now Let's revert back to question

= ( cos^2x + sin^2x )( cos^2x - sin^2x) +1

= 1( cos2x) +1

= cos2x +1

= 2 cos^2x