Question

Find the value of $\dfrac{{4\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)}}{{0.4\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)}}$

Answer 1

Hint-In this question, we use the concept of trigonometric identities and also use the basic property of complex numbers. We use $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ and $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ .In this question we also use ${i^2} = - 1$ .

Complete step-by-step answer:
Given, $\dfrac{{4\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)}}{{0.4\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)}}$
We can write, $\dfrac{{10 \times 4\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)}}{{4\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)}}$
Now, first we cancel 4 from the numerator and denominator.
$\Rightarrow \dfrac{{10\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)}}{{\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)}}$
Multiply $\cos {30^0} - i\sin {30^0}$ in numerator and denominator.
$\Rightarrow \dfrac{{10\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)\left( {\cos {{30}^0} - i\sin {{30}^0}} \right)}}{{\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)\left( {\cos {{30}^0} - i\sin {{30}^0}} \right)}}$
Now, in denominator we use the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

$$\Rightarrow \dfrac{{10\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)\left( {\cos {{30}^0} - i\sin {{30}^0}} \right)}}{{{{\left( {\cos {{30}^0}} \right)}^2} - {{\left( {i\sin {{30}^0}} \right)}^2}}} \\\ \Rightarrow \dfrac{{10\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)\left( {\cos {{30}^0} - i\sin {{30}^0}} \right)}}{{{{\cos }^2}{{30}^0} - {i^2}{{\sin }^2}{{30}^0}}} \\\ \\\$$

As we know, ${i^2} = - 1$
$\Rightarrow \dfrac{{10\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)\left( {\cos {{30}^0} - i\sin {{30}^0}} \right)}}{{{{\cos }^2}{{30}^0} + {{\sin }^2}{{30}^0}}}$
We use trigonometric identity, ${\cos ^2}\theta + {\sin ^2}\theta = 1$

$$\Rightarrow \dfrac{{10\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)\left( {\cos {{30}^0} - i\sin {{30}^0}} \right)}}{1} \\\ \Rightarrow 10\left[ {\left( {\cos {{75}^0}\cos {{30}^0} + \sin {{75}^0}\sin {{30}^0}} \right) + i\left( {\sin {{75}^0}\cos {{30}^0} - \cos {{75}^0}\sin {{30}^0}} \right)} \right] \\\$$

Now, we use trigonometric identity $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ and $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ .

$$\Rightarrow 10\left[ {\cos \left( {{{75}^0} - {{30}^0}} \right) + i\sin \left( {{{75}^0} - {{30}^0}} \right)} \right] \\\ \Rightarrow 10\left( {\cos {{45}^0} + i\sin {{45}^0}} \right) \\\$$

As we know, $\cos {45^0} = \sin {45^0} = \dfrac{1}{{\sqrt 2 }}$

$$\Rightarrow 10\left( {\dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right) \\\ \Rightarrow \dfrac{{10}}{{\sqrt 2 }}\left( {1 + i} \right) \\\ \Rightarrow 5\sqrt 2 \left( {1 + i} \right) \\\$$

So, the value of $\dfrac{{4\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)}}{{0.4\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)}}$ is $5\sqrt 2 \left( {1 + i} \right)$

Note-In such types of problems we can use two different methods. First method we already mention in above and in second method, we use the Euler’s formula ${e^{i\theta }} = \cos \theta + i\sin \theta$ to solve questions in an easy way. First we convert trigonometric terms into Euler’s form and solve them. Then we get a specific angle and express Euler’s form into a trigonometric term.
$\dfrac{{4\left( {\cos {{75}^0} + i\sin {{75}^0}} \right)}}{{0.4\left( {\cos {{30}^0} + i\sin {{30}^0}} \right)}} \Rightarrow \dfrac{{10{e^{i{{75}^0}}}}}{{{e^{i{{30}^0}}}}} \Rightarrow 10{e^{i{{45}^0}}}$
Now, express Euler’s form into a trigonometric term.
$\Rightarrow 10\left( {\cos {{45}^0} + i\sin {{45}^0}} \right) \\\ \Rightarrow 10\left( {\dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right) \\\ \Rightarrow \dfrac{{10}}{{\sqrt 2 }}\left( {1 + i} \right) \\\ \Rightarrow 5\sqrt 2 \left( {1 + i} \right) \\\$