# Find the value of alpha and theta?

## ${Z}_{1} = 2 \left(\cos \left({18}^{\circ}\right) + i \sin \left({18}^{\circ}\right)\right)$ ${Z}_{2} = \alpha \left(\cos \left({40}^{\circ}\right) + i \sin \left({40}^{\circ}\right)\right)$ ${Z}_{3} = 32 \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$ Find value of $\alpha$ and $\theta$ if $\frac{{Z}_{1}^{10}}{{Z}_{2} - {Z}_{3}} = 4 i$

Jan 19, 2018

Given:

${Z}_{1} = 2 \left(\cos \left({18}^{\circ}\right) + i \sin \left({18}^{\circ}\right)\right) \text{ [1]}$

${Z}_{2} = \alpha \left(\cos \left({40}^{\circ}\right) + i \sin \left({40}^{\circ}\right)\right) \text{ [2]}$

${Z}_{3} = 32 \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right) \text{ [3]}$

$\frac{{Z}_{1}^{10}}{{Z}_{2} - {Z}_{3}} = 4 i \text{ [4]}$

Substitute equation [1] into equation [4]:

$\frac{{\left(2 \left(\cos \left({18}^{\circ}\right) + i \sin \left({18}^{\circ}\right)\right)\right)}^{10}}{{Z}_{2} - {Z}_{3}} = 4 i$

Use De Moivre's formula:

$\frac{{2}^{10} \left(\cos \left({18}^{\circ} \times 10\right) + i \sin \left({18}^{\circ} \times 10\right)\right)}{{Z}_{2} - {Z}_{3}} = 4 i$

$\frac{1024 \left(\cos \left({180}^{\circ}\right) + i \sin \left({180}^{\circ}\right)\right)}{{Z}_{2} - {Z}_{3}} = 4 i$

$\frac{1024 \left(- 1 + i 0\right)}{{Z}_{2} - {Z}_{3}} = 4 i$

$\frac{- 1024 + i 0}{{Z}_{2} - {Z}_{3}} = 4 i \text{ [4.1]}$

Multiply both sides of equation [4.1] by ${Z}_{3} - {Z}_{2}$:

$1024 + i 0 = 4 i \left({Z}_{3} - {Z}_{2}\right) \text{ [4.2]}$

Substitute equations [2] and [3] into equation [4.2]:

$1024 + i 0 = 4 i \left(32 \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right) - \alpha \left(\cos \left({40}^{\circ}\right) + i \sin \left({40}^{\circ}\right)\right)\right)$

Perform the multiplication:

1024+i0=4i(32cos(theta)+i32sin(theta))-alphacos(40^@)-ialphasin(40^@))

1024+i0=i128cos(theta)-128sin(theta)-i4alphacos(40^@)+4alphasin(40^@))

Separating the real and imaginary parts into two equations:

$1024 = 4 \alpha \sin \left({40}^{\circ}\right) - 128 \sin \left(\theta\right) \text{ [5]}$

$i 0 = i 128 \cos \left(\theta\right) - i 4 \alpha \cos \left({40}^{\circ}\right) \text{ [6]}$

Solve equation [6] for $\alpha$ in terms of $\theta$:

$0 = 32 \cos \left(\theta\right) - \alpha \cos \left({40}^{\circ}\right)$

$\alpha = 32 \cos \frac{\theta}{\cos} \left({40}^{\circ}\right) \text{ [6.1]}$

Substitute equation [6.1] into equation [5]:

$1024 = 4 \left(32 \cos \frac{\theta}{\cos} \left({40}^{\circ}\right)\right) \sin \left({40}^{\circ}\right) - 128 \sin \left(\theta\right)$

Multiply both sides by $\cos \frac{{40}^{\circ}}{128}$:

$8 \cos \left({40}^{\circ}\right) = \cos \left(\theta\right) \sin \left({40}^{\circ}\right) - \sin \left(\theta\right) \cos \left({40}^{\circ}\right)$

Multiply both sides by -1:

$- 8 \cos \left({40}^{\circ}\right) = \sin \left(\theta\right) \cos \left({40}^{\circ}\right) - \cos \left(\theta\right) \sin \left({40}^{\circ}\right)$

The right side is the identity for $\sin \left(A - B\right)$ where $A = \theta$ and $B = {40}^{\circ}$:

$\sin \left(\theta - {40}^{\circ}\right) = - 8 \cos \left({40}^{\circ}\right)$

The above equation does not have a solution, because the right side is outside of the the range of the sine function, $- 1 \le \sin \left(x\right) \le 1$. This means that there are not real values for $\theta$ and $\alpha$