
How do you write the trigonometric form into a complex number in standard form $6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$?
Answer
558.6k+ views
Hint: This problem deals with converting the given trigonometric form into a complex number of a standard form. Here Euler’s formula is used, which is given by, if there is a complex number in the form
\[z = r\left( {\cos \theta + i\sin \theta } \right)\] which is written in polar form, then by using the Euler’s formula to write it even more concisely in exponential form : $z = r{e^{i\theta }}$.
Complete step-by-step answer:
Here the given expression is in trigonometric expression which is $6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$.
Consider the given expression below, as shown below:
$ \Rightarrow 6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$
Here on comparing the given trigonometric expression with standard form \[r\left( {\cos \theta + i\sin \theta } \right)\], we can conclude that the angle $\theta = \dfrac{{5\pi }}{{12}}$ and the radius $r = 6$.
So the expression of the complex number form $6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$ which is in the polar form by using the Euler’s formula, it can be written into the exponential form as $r{e^{i\theta }}$.
$ \Rightarrow 6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right) = 6{e^{i\dfrac{{5\pi }}{{12}}}}$
So the given expression which is in the polar form can be written into the exponential form, as shown in the above case.
Here in any case, if the complex number is written such that it is expressed in a complex number form where the real part of the complex number is equal to the cosine of an angle and the imaginary part of the complex number is the sine of the same angle, then it can be expressed in the exponential form by using the Euler’s formula.
Final answer: $\therefore 6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right) = 6{e^{i\dfrac{{5\pi }}{{12}}}}$
Note:
Please note that the Euler’s formula is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function.
\[z = r\left( {\cos \theta + i\sin \theta } \right)\] which is written in polar form, then by using the Euler’s formula to write it even more concisely in exponential form : $z = r{e^{i\theta }}$.
Complete step-by-step answer:
Here the given expression is in trigonometric expression which is $6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$.
Consider the given expression below, as shown below:
$ \Rightarrow 6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$
Here on comparing the given trigonometric expression with standard form \[r\left( {\cos \theta + i\sin \theta } \right)\], we can conclude that the angle $\theta = \dfrac{{5\pi }}{{12}}$ and the radius $r = 6$.
So the expression of the complex number form $6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right)$ which is in the polar form by using the Euler’s formula, it can be written into the exponential form as $r{e^{i\theta }}$.
$ \Rightarrow 6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right) = 6{e^{i\dfrac{{5\pi }}{{12}}}}$
So the given expression which is in the polar form can be written into the exponential form, as shown in the above case.
Here in any case, if the complex number is written such that it is expressed in a complex number form where the real part of the complex number is equal to the cosine of an angle and the imaginary part of the complex number is the sine of the same angle, then it can be expressed in the exponential form by using the Euler’s formula.
Final answer: $\therefore 6\left( {\cos \left( {\dfrac{{5\pi }}{{12}}} \right) + i\sin \left( {\dfrac{{5\pi }}{{12}}} \right)} \right) = 6{e^{i\dfrac{{5\pi }}{{12}}}}$
Note:
Please note that the Euler’s formula is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function.
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