Question

How do you perform the operation in trigonometric form $\left( \dfrac{5}{3}\left( \cos \left( 140 \right)+i\sin \left( 140 \right) \right) \right)\left( \dfrac{2}{3}\left( \cos \left( 60 \right)+i\sin \left( 60 \right) \right) \right)$?

Answer 1

This question is from the topic of algebra. In this question, we will find out the solved value of the term given in the question. In solving this question, we will understand about the term iota. After that, we will understand about the foil method formula. After solving the given term, we will see some trigonometric formulas and use them to solve the further question and get our answer.

Complete step-by-step answer:
Let us solve this question.
In this question, we have asked to solve the term $\left( \dfrac{5}{3}\left( \cos \left( 140 \right)+i\sin \left( 140 \right) \right) \right)\left( \dfrac{2}{3}\left( \cos \left( 60 \right)+i\sin \left( 60 \right) \right) \right)$.
Let us first understand about the term iota. The term iota is an imaginary number. It is square root of negative of 1. We can write iota as
$i=\sqrt{-1}$
${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$
${{i}^{3}}=-i$
${{i}^{4}}=1$
Now, let us know about the foil method. The foil method says that $\left( a+b \right)\left( c+d \right)$ can also be written as $ac+ad+bc+bd$. Or, we can write the foil method formula as
$\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd$
Now, let us solve the term $\left( \dfrac{5}{3}\left( \cos \left( 140 \right)+i\sin \left( 140 \right) \right) \right)\left( \dfrac{2}{3}\left( \cos \left( 60 \right)+i\sin \left( 60 \right) \right) \right)$. This term can also be written as
$\Rightarrow \dfrac{5}{3}\times \dfrac{2}{3}\left( \cos \left( 140 \right)+i\sin \left( 140 \right) \right)\left( \cos \left( 60 \right)+i\sin \left( 60 \right) \right)$
The above can also be written as
$\Rightarrow \dfrac{10}{9}\left( \cos \left( 140 \right)+i\sin \left( 140 \right) \right)\left( \cos \left( 60 \right)+i\sin \left( 60 \right) \right)$
So, using foil method formula, we can write the above term as
$\Rightarrow \dfrac{10}{9}\left( \cos \left( 140 \right)\cos \left( 60 \right)+\cos \left( 140 \right)i\sin \left( 60 \right)+i\sin \left( 140 \right)\cos \left( 60 \right)+i\sin \left( 140 \right)i\sin \left( 60 \right) \right)$
The above can also be written as
$\Rightarrow \dfrac{10}{9}\left( \cos \left( 140 \right)\cos \left( 60 \right)+i\cos \left( 140 \right)\sin \left( 60 \right)+i\sin \left( 140 \right)\cos \left( 60 \right)+{{i}^{2}}\sin \left( 140 \right)\sin \left( 60 \right) \right)$
Using the formula: ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$, we can write the above as
$\Rightarrow \dfrac{10}{9}\left( \cos \left( 140 \right)\cos \left( 60 \right)+i\cos \left( 140 \right)\sin \left( 60 \right)+i\sin \left( 140 \right)\cos \left( 60 \right)-\sin \left( 140 \right)\sin \left( 60 \right) \right)$
$\Rightarrow \dfrac{10}{9}\left( \left[ \cos \left( 140 \right)\cos \left( 60 \right)-\sin \left( 140 \right)\sin \left( 60 \right) \right]+i\left[ \cos \left( 140 \right)\sin \left( 60 \right)+\sin \left( 140 \right)\cos \left( 60 \right) \right] \right)$
Using the formulas: $\cos x\cos y-\sin x\sin y=\cos \left( x+y \right)$ and $\sin x\cos y+\cos x\sin y=\sin \left( x+y \right)$, we can write the above term as
$\Rightarrow \dfrac{10}{9}\left( \left[ \cos \left( 140+60 \right) \right]+i\left[ \sin \left( 60+140 \right) \right] \right)$
The above term can also be written as
$\Rightarrow \dfrac{10}{9}\left( \cos \left( 200 \right)+i\sin \left( 200 \right) \right)$
Now, we have performed the operation in trigonometric form to solve the term $\left( \dfrac{5}{3}\left( \cos \left( 140 \right)+i\sin \left( 140 \right) \right) \right)\left( \dfrac{2}{3}\left( \cos \left( 60 \right)+i\sin \left( 60 \right) \right) \right)$. The solved term is $\dfrac{10}{9}\left( \cos \left( 200 \right)+i\sin \left( 200 \right) \right)$.

Note: We should have a better knowledge in the topic of trigonometry to solve this type of question easily. Remember that all the terms inside sin and cos are in degrees. Remember the following formulas to solve this type of question easily: