Can you explain to me from where it came equation No.3?

Question

Equation NO.1 $$ g_k = \hat g \cos(\omega t_k + \varphi)$$

Equation NO.2 $$ g_{k-1} = \hat g \cos(\omega (t_k - \delta t) + \varphi)$$

Can you show me how did we get Equation No. 3?

$$ g_{k-1} =g_k \cos(\omega \delta t) + \hat g\sin(\omega t_k + \varphi)\sin(\omega \delta t)$$

Answer 1

Addition formula for cosine.

$\begin{array}\\ g_{k-1} &= \hat g \cos(\omega (t_k - \delta t) + \varphi)\\ &= \hat g \cos(\omega (t_k) + \varphi-\omega (\delta t))\\ &= \hat g \left(\cos(\omega (t_k) + \varphi)\cos(\omega \delta t)+\sin(\omega (t_k) + \varphi)\sin(\omega (\delta t)))\right)\\ &= g_k\cos(\omega \delta t)+\hat g \sin(\omega (t_k) + \varphi)\sin(\omega (\delta t)))\\ \end{array} $

Answer 2

$$g_{k-1} = \hat g \cos(\omega (t_k - \delta t) + \varphi) = \hat g \cos(\omega t_k - \omega \delta t + \varphi)$$ Using $$\cos(a-b) = \cos a \cos b + \sin a \sin b$$ where $a = \omega t_k + \varphi$ and $ b = \omega \delta t$, we get $$g_{k-1} = \underbrace{\hat g\cos (\omega t_k + \varphi)}_{g_{k}} \cos \omega \delta t+ \hat g\sin (\omega t_k + \varphi) \sin \omega \delta t$$