Solving Cos4x + Sin4x = 1: Can I Sqroot Both Terms?

[uzman1243]

I am a bit confused here.

Cos²x + sin²x = 1

Thus can I say

Cos⁴x + sin⁴x = 1

If I just sqroot each term:
sqroot Cos⁴x + sqroot sin⁴x = sqroot (1) = 1

?

 

[AdityaDev]

No. Sqrt(a+b) and sqrt(a)+sqrt(b) is different. If you take sqrt on both sides, it has to be sqrt(a+b)=sqrt(c).
But you can still write$cos^4x+sin^4x=(cos^2x+sin^2x)-2cos^2xsin^2x$

 

[PeroK]

No. Sqrt(a+b) and sqrt(a)+sqrt(b) is different. If you take sqrt on both sides, it has to be sqrt(a+b)=sqrt(c).
But you can still write$cos^4x+sin^4x=(cos^2x+sin^2x)-2cos^2xsin^2x$

You can write it, but it wouldn't be correct. Instead:

$cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2x = 1 - 2cos^2xsin^2x$

 

[Obliczone]

I think that the simplest form will be:
$$...=1-2\cos^2x\sin^2x=1-\frac{1}{2}\sin^2(2x)$$
because:
$$2\sin x\cos x=\sin(2x)$$

 

[AdityaDev]

You can write it, but it wouldn't be correct. Instead:

$cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2x = 1 - 2cos^2xsin^2x$

it was a typo

 

[Svein]

Using complex analysis makes it still more elegant. On one hand: $e^{4ix}=cos(4x)+isin(4x)$. On the other hand: $e^{4ix}=(cos(x)+isin(x))^{4}= cos^{4}(x) + 4icos^{3}(x)sin(x) + 6i^{2}cos^{2}(x)sin^{2}(x)+4i^{3}cos(x)sin^{3}(x)+i^{4}sin^{4}(x)$. Multiplying out: $e^{4ix}=cos^{4}(x) + 4icos^{3}(x)sin(x) - 6cos^{2}(x)sin^{2}(x)-4icos(x)sin^{3}(x)+sin^{4}(x)$. Thus: $cos(4x)+isin(4x)=cos^{4}(x)+sin^{4}(x) - 6cos^{2}(x)sin^{2}(x) + i(4cos^{3}(x)sin(x) )-4cos(x)sin^{3}(x))$. Taking the real part: $cos(4x)=cos^{4}(x)+sin^{4}(x) - 6cos^{2}(x)sin^{2}(x)$

 

[Byeonggon Lee]

From the definition of trigonometric function
cos=x/r
sin=y/r
From the Pythagorean theorem
r^2=x^2+y^2

cos^4(x) + sin^4(x)
=(x^4+y^4)/r^4
=((x^2+y^2)^2-2x^2y^2)/(r^4)
=(r^4-2x^2y^2)/r^4
!=1