Question

The extreme values of $4\cos \left( {{x^2}} \right)\cos \left[ {\left( {\dfrac{\pi }{3}} \right) + {x^2}} \right]\cos \left[ {\left( {\dfrac{\pi }{3}} \right) - {x^2}} \right]$ over r are
$\left( 1 \right){\text{ }} - 1,1$
$\left( 2 \right){\text{ }} - 2,2$
$\left( 3 \right){\text{ }} - 3,3$
$\left( 4 \right){\text{ }} - 4,4$

Answer 1

Try to simplify the expression by using formulas of $\cos \left( {a + b} \right)$ and $\cos \left( {a - b} \right)$ then substitute the trigonometric values when needed. After that try to identify which identity you can be used here for further solving. Then just by using formulas or identities we will be able to do them. And remember that the range of the function $\cos \theta$ is $\left[ { - 1,1} \right]$ .With the help of range you can find the extreme values.

Complete answer:
The given expression is $4\cos \left( {{x^2}} \right)\cos \left[ {\left( {\dfrac{\pi }{3}} \right) + {x^2}} \right]\cos \left[ {\left( {\dfrac{\pi }{3}} \right) - {x^2}} \right]$ .To find the extreme values, first we have to solve this expression. So, as we know that $\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$ and $\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$ .By applying these formulas in the given expression we get
$\Rightarrow 4\cos \left( {{x^2}} \right).\left[ {\cos \left( {\dfrac{\pi }{3}} \right)\cos {x^2} - \sin \left( {\dfrac{\pi }{3}} \right)\sin {x^2}} \right].\left[ {\cos \left( {\dfrac{\pi }{3}} \right)\cos {x^2} + \sin \left( {\dfrac{\pi }{3}} \right)\sin {x^2}} \right]$ ------- (i)
The value of $\cos \left( {\dfrac{\pi }{3}} \right)$ means the value of $\cos 60^\circ$ is $\dfrac{1}{2}$ .Whereas the value of $\sin \left( {\dfrac{\pi }{3}} \right)$ that is the value of $\sin 60^\circ$ is $\dfrac{{\sqrt 3 }}{2}$ . So by putting these values in the equation (i) we get
$\Rightarrow 4\cos \left( {{x^2}} \right).\left[ {\dfrac{1}{2}\cos {x^2} - \dfrac{{\sqrt 3 }}{2}\sin {x^2}} \right].\left[ {\dfrac{1}{2}\cos {x^2} + \dfrac{{\sqrt 3 }}{2}\sin {x^2}} \right]$ ------- (ii)
We also know that $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ .Now observe the equation (ii). We get that in the above equation the term $\left[ {\dfrac{1}{2}\cos {x^2} - \dfrac{{\sqrt 3 }}{2}\sin {x^2}} \right]$ is $\left( {a - b} \right)$ and the term $\left[ {\dfrac{1}{2}\cos {x^2} + \dfrac{{\sqrt 3 }}{2}\sin {x^2}} \right]$ is $\left( {a + b} \right)$ .So we can write the equation(ii) as
$\Rightarrow 4\cos \left( {{x^2}} \right).\left[ {{{\left( {\dfrac{1}{2}\cos {x^2}} \right)}^2} - {{\left( {\dfrac{{\sqrt 3 }}{2}\sin {x^2}} \right)}^2}} \right]$
And on further solving the above equation becomes
$\Rightarrow 4\cos \left( {{x^2}} \right).\left[ {\dfrac{1}{4}{{\cos }^2}{x^2} - \dfrac{3}{4}{{\sin }^2}{x^2}.} \right]$
Now take out $4$ common from the square bracket
$\Rightarrow 4\cos \left( {{x^2}} \right).\dfrac{1}{4}\left[ {{{\cos }^2}{x^2} - 3{{\sin }^2}{x^2}.} \right]$
The $4$ in the numerator and the $4$ in the denominator will cancel out each other and we are left with
$\Rightarrow \cos \left( {{x^2}} \right).\left[ {{{\cos }^2}{x^2} - 3{{\sin }^2}{x^2}.} \right]$ --------- (iii)
We know that ${\cos ^2}x + {\sin ^2}x = 1$ so we can write ${\sin ^2}x$ as $1 - {\cos ^2}x$ .Hence the equation (iii) becomes
$\Rightarrow \cos \left( {{x^2}} \right).\left[ {{{\cos }^2}{x^2} - 3\left( {1 - {{\cos }^2}{x^2}} \right).} \right]$
Open the bracket by multiplying with $3$
$\Rightarrow \cos \left( {{x^2}} \right).\left[ {{{\cos }^2}{x^2} - 3 + 3{{\cos }^2}{x^2}} \right]$
Add the similar terms inside the bracket
$\Rightarrow \cos \left( {{x^2}} \right).\left[ {4{{\cos }^2}{x^2} - 3} \right]$
On further solving the above equation becomes
$\Rightarrow 4{\cos ^3}{x^2} - 3\cos {x^2}$ ----- (iv)
The above value is similar to the formula of cos of three times of $\theta$ which is given by $\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$ .So our equation (iv) becomes
$\Rightarrow \cos \left( {3{x^2}} \right)$
As the value of $\cos \theta$ lies between $- 1$ and $1$ that is $- 1 \leqslant \cos \theta \leqslant 1$ .So $\cos 3\theta$ can also lies between $- 1$ and $1$ .Here we have $\theta = {x^2}$ .Therefore we can write it as
$\Rightarrow - 1 \leqslant \cos 3{x^2} \leqslant 1$
Hence the required extreme values are $- 1$ and $1$ .
Thus, the correct option is $\left( 1 \right){\text{ }} - 1,1$ .

Note:
You have to first learn all the formulas of trigonometry. You should know all the formulas so that you will be able to attend any question. Like in this question we use the formulas $\cos \left( {a + b} \right)$ and $\cos \left( {a - b} \right)$ in the first step of out answer so there may be a chance of a mistake as both the formulas are similar. So learn, use and write them carefully.