If ${e^{i\theta }} = \cos \theta + i\sin \theta $ then in \[\Delta ABC\] value of ${e^{iA}}.{e^{iB}}.{e^{iC}}$ is
A) \[ - i\]
B) \[1\]
C) \[ - 1\]
D) None of these
Answer
527.4k+ views
Hint: In order to find the value of ${e^{iA}}.{e^{iB}}.{e^{iC}}$, initiate with expressing every term of the value into its corresponding complex number representation. Using the basic triangle property that is the sum of interior angles of a triangle, then apply the law of radicals to get the sum of powers and hence substitute the value of the required trigonometric function.
Complete answer: We are given with an information that ${e^{i\theta }} = \cos \theta + i\sin \theta $, and we need to find the value of ${e^{iA}}.{e^{iB}}.{e^{iC}}$ for a triangle ABC.
For that, we know in a triangle, the sum of interior angles of a triangle is always equal to \[{180^ \circ }\] i.e,\[A + B + C = {180^ \circ }\]
So, In ABC, \[A + B + C = {180^ \circ } = \pi \] …….(1)
Therefore, in \[{e^{iA}}.{e^{iB}}.{e^{iC}}\]:
We can see that the base value is same for all, so according to the Law of Radicals, the powers will be added, so we get:
\[ = {e^{iA + iB + iC}}\]
Taking i common in the power:
\[ = {e^{i\left( {A + B + C} \right)}}\]
Substituting the value of equation 1 in the above equation, and we get:
\[ \Rightarrow {e^{i\left( {A + B + C} \right)}} = {e^{i\pi }}\]
Since, we were given that ${e^{i\theta }} = \cos \theta + i\sin \theta $, so comparing \[{e^{i\theta }}\] with \[{e^{i\pi }}\], we get:
\[\theta = \pi \]
So, substituting \[\theta = \pi \] in ${e^{i\theta }} = \cos \theta + i\sin \theta $ ,we get:
${e^{i\pi }} = \cos \pi + i\sin \pi $
Since, we know that:
$\sin \pi = 0$ and $\cos \pi = - 1$
So, substituting the value in equation 2, we get:
${e^{i\pi }} = - 1 + i \times 0$
$ \Rightarrow {e^{i\pi }} = - 1$
Therefore, Option (C) is the correct answer.
Note:
1.The equation given to us ${e^{i\theta }} = \cos \theta + i\sin \theta $, is also known as the Euler’s formula, which shows the fundamental relation between the trigonometric function and complex exponential function.
2.According to the Law of radicals, if the base is the same in a product then their powers will be added. For example: In ${p^a}.{p^b}$, the base is common that is p, so the powers are added, and we get: ${p^a}.{p^b} = {p^{a + b}}$.
Complete answer: We are given with an information that ${e^{i\theta }} = \cos \theta + i\sin \theta $, and we need to find the value of ${e^{iA}}.{e^{iB}}.{e^{iC}}$ for a triangle ABC.
For that, we know in a triangle, the sum of interior angles of a triangle is always equal to \[{180^ \circ }\] i.e,\[A + B + C = {180^ \circ }\]
So, In ABC, \[A + B + C = {180^ \circ } = \pi \] …….(1)
Therefore, in \[{e^{iA}}.{e^{iB}}.{e^{iC}}\]:
We can see that the base value is same for all, so according to the Law of Radicals, the powers will be added, so we get:
\[ = {e^{iA + iB + iC}}\]
Taking i common in the power:
\[ = {e^{i\left( {A + B + C} \right)}}\]
Substituting the value of equation 1 in the above equation, and we get:
\[ \Rightarrow {e^{i\left( {A + B + C} \right)}} = {e^{i\pi }}\]
Since, we were given that ${e^{i\theta }} = \cos \theta + i\sin \theta $, so comparing \[{e^{i\theta }}\] with \[{e^{i\pi }}\], we get:
\[\theta = \pi \]
So, substituting \[\theta = \pi \] in ${e^{i\theta }} = \cos \theta + i\sin \theta $ ,we get:
${e^{i\pi }} = \cos \pi + i\sin \pi $
Since, we know that:
$\sin \pi = 0$ and $\cos \pi = - 1$
So, substituting the value in equation 2, we get:
${e^{i\pi }} = - 1 + i \times 0$
$ \Rightarrow {e^{i\pi }} = - 1$
Therefore, Option (C) is the correct answer.
Note:
1.The equation given to us ${e^{i\theta }} = \cos \theta + i\sin \theta $, is also known as the Euler’s formula, which shows the fundamental relation between the trigonometric function and complex exponential function.
2.According to the Law of radicals, if the base is the same in a product then their powers will be added. For example: In ${p^a}.{p^b}$, the base is common that is p, so the powers are added, and we get: ${p^a}.{p^b} = {p^{a + b}}$.
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