
Let \[f\left( \theta \right) = \left| {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\cos \theta \sin \theta }&{ - \sin \theta }\\{\cos \theta \sin \theta }&{{{\sin }^2}\theta }&{\cos \theta }\\{\sin \theta }&{ - \cos \theta }&0\end{array}} \right|\] then \[f\left( {\dfrac{\pi }{6}} \right) = \]
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Answer
566.4k+ views
Hint: In the given question, we have been given a function which is to be evaluated at a particular given value of the argument (here, it is ‘angle’ which is being represented by \[\theta \]). The function contains some terms which are a combination of two trigonometric functions – sine, cosine. These functions are in a determinant whose value is to be calculated at the given value. So, we are going to approach this question by first solving the determinant part of the function by applying the required formula and using some basic identities to simplify it. Then, after completely calculating the answer, we are going to put in the given value of the argument and that is going to give us our answer.
Formula Used:
In the given question we are going to put in the formula of determinant, which is:
\[\left| {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right| = a\left( {e \times i - h \times f} \right) - b\left( {d \times i - g \times f} \right) + c\left( {d \times h - g \times e} \right)\]
Complete step-by-step answer:
The given expression in the question is \[f\left( \theta \right) = \left| {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\cos \theta \sin \theta }&{ - \sin \theta }\\{\cos \theta \sin \theta }&{{{\sin }^2}\theta }&{\cos \theta }\\{\sin \theta }&{ - \cos \theta }&0\end{array}} \right|\].
First, we are going to solve the determinant by applying the formula:
\[\left| {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\cos \theta \sin \theta }&{ - \sin \theta }\\{\cos \theta \sin \theta }&{{{\sin }^2}\theta }&{\cos \theta }\\{\sin \theta }&{ - \cos \theta }&0\end{array}} \right|\]
\[ = {\cos ^2}\theta \left( {{{\sin }^2}\theta \times 0 - \left( { - \cos \theta \times \cos \theta } \right)} \right) - \cos \theta \sin \theta \left( {\cos \theta \sin \theta \times 0 - \sin \theta \times \cos \theta } \right) + \left( { - \sin \theta } \right)\left( {\cos \theta \sin \theta \times \left( { - \cos \theta } \right) - \sin \theta \times {{\sin }^2}\theta } \right)\]
Simplifying the brackets and solving,
\[ = {\cos ^4}\theta + {\cos ^2}\theta {\sin ^2}\theta + {\cos ^2}\theta {\sin ^2}\theta + {\sin ^4}\theta \]
\[ = {\cos ^4}\theta + {\sin ^4}\theta + 2{\cos ^2}\theta {\sin ^2}\theta \]
Clearly, this is similar to the identity of \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[ = {\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)^2} = {\left( 1 \right)^2} = 1\]
Hence, the answer of the given expression is \[1\].
Hence, the correct option is (2).
Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contains the known and the unknown and pick the one which is the most suitable for the answer. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.
Formula Used:
In the given question we are going to put in the formula of determinant, which is:
\[\left| {\begin{array}{*{20}{c}}a&b&c\\d&e&f\\g&h&i\end{array}} \right| = a\left( {e \times i - h \times f} \right) - b\left( {d \times i - g \times f} \right) + c\left( {d \times h - g \times e} \right)\]
Complete step-by-step answer:
The given expression in the question is \[f\left( \theta \right) = \left| {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\cos \theta \sin \theta }&{ - \sin \theta }\\{\cos \theta \sin \theta }&{{{\sin }^2}\theta }&{\cos \theta }\\{\sin \theta }&{ - \cos \theta }&0\end{array}} \right|\].
First, we are going to solve the determinant by applying the formula:
\[\left| {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\cos \theta \sin \theta }&{ - \sin \theta }\\{\cos \theta \sin \theta }&{{{\sin }^2}\theta }&{\cos \theta }\\{\sin \theta }&{ - \cos \theta }&0\end{array}} \right|\]
\[ = {\cos ^2}\theta \left( {{{\sin }^2}\theta \times 0 - \left( { - \cos \theta \times \cos \theta } \right)} \right) - \cos \theta \sin \theta \left( {\cos \theta \sin \theta \times 0 - \sin \theta \times \cos \theta } \right) + \left( { - \sin \theta } \right)\left( {\cos \theta \sin \theta \times \left( { - \cos \theta } \right) - \sin \theta \times {{\sin }^2}\theta } \right)\]
Simplifying the brackets and solving,
\[ = {\cos ^4}\theta + {\cos ^2}\theta {\sin ^2}\theta + {\cos ^2}\theta {\sin ^2}\theta + {\sin ^4}\theta \]
\[ = {\cos ^4}\theta + {\sin ^4}\theta + 2{\cos ^2}\theta {\sin ^2}\theta \]
Clearly, this is similar to the identity of \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[ = {\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)^2} = {\left( 1 \right)^2} = 1\]
Hence, the answer of the given expression is \[1\].
Hence, the correct option is (2).
Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contains the known and the unknown and pick the one which is the most suitable for the answer. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.
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