If $\omega $ is an imaginary cube root of unity, then the value of $\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right]$ is
$\left( a \right)\dfrac{{ - \sqrt 3 }}{2}$
\[\left( b \right)\dfrac{{ - 1}}{{\sqrt 2 }}\]
$\left( c \right)\dfrac{1}{{\sqrt 2 }}$
$\left( d \right)\dfrac{{\sqrt 3 }}{2}$
Answer
661.2k+ views
Hint: In this particular type of question use the concept that if $\omega $ is an imaginary cube root of unity then (\[{\omega ^3} = 1\]) and (${\omega ^2} + \omega + 1 = 0$) and use the concept that sin (-x) = -sin x and sine is negative in the third quadrant. So use these concepts to reach the solution of the question.
Complete step-by-step answer:
As it is given that $\omega $ is an imaginary cube root of unity, so it satisfies the following condition which is given as,
\[{\omega ^3} = 1\]......................... (1)
And
${\omega ^2} + \omega + 1 = 0$.................... (2)
Now given equation is
$\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now first simplify the above equation we have,
$ \Rightarrow \sin \left[ {\left( {{{\left( {{\omega ^3}} \right)}^3}\omega + {{\left( {{\omega ^3}} \right)}^7}{\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now substitute the value of \[{\omega ^3}\] from equation (1) in the above equation we have,
$ \Rightarrow \sin \left[ {\left( {{{\left( 1 \right)}^3}\omega + {{\left( 1 \right)}^7}{\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now as we know any power of 1 is always remain 1 so use this property to simplify the above equation we have,
\[ \Rightarrow \sin \left[ {\left( {\omega + {\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]\].......................... (3)
Now from equation (2) we have,
$ \Rightarrow {\omega ^2} + \omega + 1 = 0$
$ \Rightarrow {\omega ^2} + \omega = - 1$
So substitute this value in equation (3) we have,
\[ \Rightarrow \sin \left[ {\left( { - 1} \right)\pi - \dfrac{\pi }{4}} \right]\]
Now the above equation is also written as,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right]\]
Now as we know that sin (-x) = -sin x so use this property in the above equation we have,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right] = - \sin \left[ {\pi + \dfrac{\pi }{4}} \right]\]
Now as we know that sin in 3rd quadrant is negative, so ($\sin \left( {\pi + \theta } \right) = - \sin \theta $), so use this property in the above equation we have,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right] = - \sin \left[ {\pi + \dfrac{\pi }{4}} \right] = - \left( { - \sin \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4}\]
Now the given equation is convert into simple standard value of sine i.e. \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Therefore,
$\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right] = \dfrac{1}{{\sqrt 2 }}$
So this is the required answer.
Hence option (C) is the correct answer.
Note – Whenever we face such types of questions the key concept we have to remember is that always recall all the properties if $\omega $ is an imaginary cube root of unity which is all stated above, then first simplify the above equation using these properties as above, then apply the basic sine properties as above and simplify we will get the required answer.
Complete step-by-step answer:
As it is given that $\omega $ is an imaginary cube root of unity, so it satisfies the following condition which is given as,
\[{\omega ^3} = 1\]......................... (1)
And
${\omega ^2} + \omega + 1 = 0$.................... (2)
Now given equation is
$\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now first simplify the above equation we have,
$ \Rightarrow \sin \left[ {\left( {{{\left( {{\omega ^3}} \right)}^3}\omega + {{\left( {{\omega ^3}} \right)}^7}{\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now substitute the value of \[{\omega ^3}\] from equation (1) in the above equation we have,
$ \Rightarrow \sin \left[ {\left( {{{\left( 1 \right)}^3}\omega + {{\left( 1 \right)}^7}{\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]$
Now as we know any power of 1 is always remain 1 so use this property to simplify the above equation we have,
\[ \Rightarrow \sin \left[ {\left( {\omega + {\omega ^2}} \right)\pi - \dfrac{\pi }{4}} \right]\].......................... (3)
Now from equation (2) we have,
$ \Rightarrow {\omega ^2} + \omega + 1 = 0$
$ \Rightarrow {\omega ^2} + \omega = - 1$
So substitute this value in equation (3) we have,
\[ \Rightarrow \sin \left[ {\left( { - 1} \right)\pi - \dfrac{\pi }{4}} \right]\]
Now the above equation is also written as,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right]\]
Now as we know that sin (-x) = -sin x so use this property in the above equation we have,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right] = - \sin \left[ {\pi + \dfrac{\pi }{4}} \right]\]
Now as we know that sin in 3rd quadrant is negative, so ($\sin \left( {\pi + \theta } \right) = - \sin \theta $), so use this property in the above equation we have,
\[ \Rightarrow \sin \left[ { - \left( {\pi + \dfrac{\pi }{4}} \right)} \right] = - \sin \left[ {\pi + \dfrac{\pi }{4}} \right] = - \left( { - \sin \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4}\]
Now the given equation is convert into simple standard value of sine i.e. \[\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
Therefore,
$\sin \left[ {\left( {{\omega ^{10}} + {\omega ^{23}}} \right)\pi - \dfrac{\pi }{4}} \right] = \dfrac{1}{{\sqrt 2 }}$
So this is the required answer.
Hence option (C) is the correct answer.
Note – Whenever we face such types of questions the key concept we have to remember is that always recall all the properties if $\omega $ is an imaginary cube root of unity which is all stated above, then first simplify the above equation using these properties as above, then apply the basic sine properties as above and simplify we will get the required answer.
Recently Updated Pages
Explain the structure of megasporangium class 12 biology CBSE

Why is chloroform kept in dark coloured bottles class 12 chemistry CBSE

Derive the balancing condition of a Wheatstone bri class 12 physics CBSE

Draw VI characteristics of a pn junction diode Explain class 12 physics CBSE

Is it possible to carry pride in ones language too class 12 english CBSE

In a population at Hardy Weinberg equilibrium the allele class 12 biology CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

