 # The equation is a cos &#x2061;<!-- ⁡ --> ( &#x03C9;<!-- ω --> t Aedan Gonzales 2022-05-07 Answered
The equation is
$a\mathrm{cos}\left(\omega t\right)+b\mathrm{sin}\left(\omega t\right)=A\mathrm{cos}\left(\omega t+\varphi \right)$
I must have forgotten how the right hand side follows from the left hand side.
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Let $A=\sqrt{{a}^{2}+{b}^{2}}$ and $\varphi$ be an angle such that $\mathrm{cos}\varphi =\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}$ and $\mathrm{sin}\varphi =\frac{-b}{\sqrt{{a}^{2}+{b}^{2}}}$
Then, by using the identity $\mathrm{cos}\left(x+y\right)=\mathrm{cos}x\mathrm{cos}y-\mathrm{sin}x\mathrm{sin}y$, we have:
$\begin{array}{rl}a\mathrm{cos}\left(\omega t\right)+b\mathrm{sin}\left(\omega t\right)& =\sqrt{{a}^{2}+{b}^{2}}\left[\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}\mathrm{cos}\left(\omega t\right)-\frac{-b}{\sqrt{{a}^{2}+{b}^{2}}}\mathrm{sin}\left(\omega t\right)\right]\\ & =A\left[\mathrm{cos}\varphi \mathrm{cos}\left(\omega t\right)-\mathrm{sin}\varphi \mathrm{sin}\left(\omega t\right)\right]\\ & =A\mathrm{cos}\left(\omega t+\varphi \right).\end{array}$
###### Not exactly what you’re looking for? Hailee Stout
Try breaking down $A\mathrm{cos}\left(\omega t+\varphi \right)$
$\begin{array}{lll}A\mathrm{cos}\left(\omega t+\varphi \right)& =& A\left(\mathrm{cos}\omega t\mathrm{cos}\varphi -\mathrm{sin}\omega t\mathrm{sin}\varphi \right)\\ & =& \left(A\mathrm{cos}\varphi \right)\mathrm{cos}\omega t-\left(A\mathrm{sin}\varphi \right)\mathrm{sin}\omega t\end{array}$
Try to show that for arbitrary constants a and b that there exists constants A and $\varphi$ such that
$a=A\mathrm{cos}\varphi$
and
$b=A\mathrm{sin}\varphi$
What is ${a}^{2}+{b}^{2}$?