Courtney Griffin

2022-01-26

Prove that ${\mathrm{cos}}^{4}\left(\theta \right)-{\mathrm{sin}}^{4}\left(\theta \right)=\mathrm{cos}\left(2\theta \right)$
So, here are my steps so far:
${\left({\mathrm{cos}}^{2}\theta \right)}^{2}-{\left({\mathrm{sin}}^{2}\theta \right)}^{2}=\mathrm{cos}\left(2\theta \right)$
$\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)=\mathrm{cos}\left(2\theta \right)$
$\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)=\mathrm{cos}\left(2\theta \right)$

nebajcioz

Expert

Well, ${\mathrm{cos}}^{2}\left(\theta \right)+{\mathrm{sin}}^{2}\left(\theta \right)=1$, and $\mathrm{cos}\left(2\theta \right)={\mathrm{cos}}^{2}\left(\theta \right)-{\mathrm{sin}}^{2}\left(\theta \right)$ is a standard Double-Angle identity.

Prince Huang

Expert

We have
${\mathrm{cos}}^{4}\left(\theta \right)-{\mathrm{sin}}^{4}\left(\theta \right)=\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)$
where $\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)=1$ and $\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)=2{\mathrm{cos}}^{2}\left(\theta \right)-1=\mathrm{cos}\left(2\theta \right)$

2022-01-31

No the answer is /cos sintheta 7

Do you have a similar question?