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

Question

How does one instantly know that

$$(1-\cos(2x))^2$$

leads to

$$(\cos^2(x) + \sin^2(x)-(\cos^2(x)-\sin^2(x))^2$$

I know that

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

but it confuses me that I can't get to what's written above.

I got it from here (see denominator):

@an4s Ah now I see. Thanks :) – Ramanujan Taylor Mar 04 '20 at 16:55 — Ramanujan Taylor, Mar 04 '20 at 16:55

score 1 · Answer 1 · answered Mar 04 '20 at 17:13

1

$$(\cos^2x + \sin^2x-(\cos^2x-\sin^2x))^2=(1-\cos2x)^2$$ $$\cos^2x + \sin^2x=1$$ and $$\cos^2x -\sin^2x=\cos2x.$$

answered Mar 04 '20 at 17:13

Michael Rozenberg

Matteo · Accepted Answer · 2020-03-04T18:41:03.227

1

It's very simple, in fact: $\sin^2(x)+\cos^2(x)=1$ and $\cos(2x)=\cos^2(x)-\sin^2(x)$

So, substituting, I obtain: $$(1-\cos(2x))^2=(\cos^2(x) + \sin^2(x)-(\sin^2(x)-\cos^2(x))^2=(2\cos^2(x))^2=4\cos^4(x)$$

edited Mar 04 '20 at 18:41

answered Mar 04 '20 at 17:19

Matteo

I believe it's actually $\cos(2x)=\cos^2x-\sin^2x$... – Dr. Mathva Mar 04 '20 at 18:37

2 Answers2