Question

$$\left. \begin{array} { l } { \cos e } \\ { 2 \sin A = 2 \cos A } \end{array} \right.$$

Answer

a=(1-cosA/2)/(sinA*n*d)

Solution


Regroup terms.
\[2sinAand=2-cosA\]
Divide both sides by \(2\).
\[sinAand=\frac{2-cosA}{2}\]
Simplify  \(\frac{2-cosA}{2}\)  to  \(1-\frac{cosA}{2}\).
\[sinAand=1-\frac{cosA}{2}\]
Divide both sides by \(sinA\).
\[and=\frac{1-\frac{cosA}{2}}{sinA}\]
Divide both sides by \(n\).
\[ad=\frac{\frac{1-\frac{cosA}{2}}{sinA}}{n}\]
Simplify  \(\frac{\frac{1-\frac{cosA}{2}}{sinA}}{n}\)  to  \(\frac{1-\frac{cosA}{2}}{sinAn}\).
\[ad=\frac{1-\frac{cosA}{2}}{sinAn}\]
Divide both sides by \(d\).
\[a=\frac{\frac{1-\frac{cosA}{2}}{sinAn}}{d}\]
Simplify  \(\frac{\frac{1-\frac{cosA}{2}}{sinAn}}{d}\)  to  \(\frac{1-\frac{cosA}{2}}{sinAnd}\).
\[a=\frac{1-\frac{cosA}{2}}{sinAnd}\]