
If ABC are angles of a triangle such that\[x = cisA\], \[y = cisB\], \[z = cisC\], then find value of $xyz$
(A). -1
(B). 1
(C). $ - i$
(D). $i$
Answer
216.3k+ views
Hint: Use Euler’s complex formula to write the given values in exponential complex form. Then use the supplementary condition of the interior angles of a triangle to find the value of the desired expression.
Complete step-by-step solution -

A complex number is a number that can be expressed in the form $a + ib$, where $a$ and $b$ are real numbers.
The acronym $cis$ is a commonly used mathematical notation for complex numbers defined by:
$cis\left( x \right) = \cos \left( x \right) + i\sin \left( x \right)$
Given the problem, ABC is a triangle.
The angles of the triangle are given as:
$
x = cisA \\
y = cisB \\
z = cisC \\
$
We need to find the value of the expression $xyz$.
Euler’s formula on the complex numbers states that ${e^{i\theta }} = \left( {\cos \theta + i\sin \theta } \right)$.
Using the same in above given data, we get:
$\left(
x = cisA = \cos A + i\sin A = {e^{iA}} \\
y = cisB = \cos B + i\sin B = {e^{iB}} \\
z = cisC = \cos C + i\sin C = {e^{iC}} \\
\right){\text{ (1)}}$
Also, we know that for a triangle, the sum of all its interior angles is supplementary or ${180^0}$.
Using the same result for triangle ABC, we get
$A + B + C = {180^0}{\text{ (2)}}$
We need to find the value of the expression $xyz$.
Using equations (1) in expression$xyz$, we get
$
xyz = cisA.cisB.cisC \\
\Rightarrow xyz = \left( {\cos A + i\sin A} \right)\left( {\cos B + i\sin B} \right)\left( {\cos C + i\sin C} \right) \\
\Rightarrow xyz = {e^{iA}}.{e^{iB}}.{e^{iC}}{\text{ (3)}} \\
$
We know that the product of two exponentials with same base is given by
${a^b} \times {a^c} = {a^{b + c}}$
Using this result in equation (3), we get
$ \Rightarrow xyz = {e^{iA}}.{e^{iB}}.{e^{iC}} = {e^{i\left( {A + B + C} \right)}}$
Using equation (2) in above, we get
$ \Rightarrow xyz = {e^{i\left( {A + B + C} \right)}} = {e^{i\left( {{{180}^0}} \right)}}$
Transforming the above equation using Euler’s formula for complex number as stated in the above part of the solution, we get
\[ \Rightarrow xyz = {e^{i\left( {{{180}^0}} \right)}} = \cos {180^0} + i\sin {180^0}\]
Using $\cos {180^0} = - 1$and $\sin {180^0} = 0$in above equation, we get
\[
\Rightarrow xyz = \cos {180^0} + i\sin {180^0} = - 1 + i \times 0 = - 1 \\
\Rightarrow xyz = - 1 \\
\]
Hence the value of the expression $xyz$ is -1.
Therefore, option (A). -1 is the correct answer.
Note: The Euler’s formula for complex numbers and the cis notation should be kept in mind while solving problems like above. Euler’s complex formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. The above formula is used for a unit complex number, that is a complex number with magnitude unity.
Complete step-by-step solution -

A complex number is a number that can be expressed in the form $a + ib$, where $a$ and $b$ are real numbers.
The acronym $cis$ is a commonly used mathematical notation for complex numbers defined by:
$cis\left( x \right) = \cos \left( x \right) + i\sin \left( x \right)$
Given the problem, ABC is a triangle.
The angles of the triangle are given as:
$
x = cisA \\
y = cisB \\
z = cisC \\
$
We need to find the value of the expression $xyz$.
Euler’s formula on the complex numbers states that ${e^{i\theta }} = \left( {\cos \theta + i\sin \theta } \right)$.
Using the same in above given data, we get:
$\left(
x = cisA = \cos A + i\sin A = {e^{iA}} \\
y = cisB = \cos B + i\sin B = {e^{iB}} \\
z = cisC = \cos C + i\sin C = {e^{iC}} \\
\right){\text{ (1)}}$
Also, we know that for a triangle, the sum of all its interior angles is supplementary or ${180^0}$.
Using the same result for triangle ABC, we get
$A + B + C = {180^0}{\text{ (2)}}$
We need to find the value of the expression $xyz$.
Using equations (1) in expression$xyz$, we get
$
xyz = cisA.cisB.cisC \\
\Rightarrow xyz = \left( {\cos A + i\sin A} \right)\left( {\cos B + i\sin B} \right)\left( {\cos C + i\sin C} \right) \\
\Rightarrow xyz = {e^{iA}}.{e^{iB}}.{e^{iC}}{\text{ (3)}} \\
$
We know that the product of two exponentials with same base is given by
${a^b} \times {a^c} = {a^{b + c}}$
Using this result in equation (3), we get
$ \Rightarrow xyz = {e^{iA}}.{e^{iB}}.{e^{iC}} = {e^{i\left( {A + B + C} \right)}}$
Using equation (2) in above, we get
$ \Rightarrow xyz = {e^{i\left( {A + B + C} \right)}} = {e^{i\left( {{{180}^0}} \right)}}$
Transforming the above equation using Euler’s formula for complex number as stated in the above part of the solution, we get
\[ \Rightarrow xyz = {e^{i\left( {{{180}^0}} \right)}} = \cos {180^0} + i\sin {180^0}\]
Using $\cos {180^0} = - 1$and $\sin {180^0} = 0$in above equation, we get
\[
\Rightarrow xyz = \cos {180^0} + i\sin {180^0} = - 1 + i \times 0 = - 1 \\
\Rightarrow xyz = - 1 \\
\]
Hence the value of the expression $xyz$ is -1.
Therefore, option (A). -1 is the correct answer.
Note: The Euler’s formula for complex numbers and the cis notation should be kept in mind while solving problems like above. Euler’s complex formula establishes the fundamental relationship between the trigonometric functions and the complex exponential function. The above formula is used for a unit complex number, that is a complex number with magnitude unity.
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