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Hint: First simplify the expression using various algebraic & trigonometric identities & then use the range of the trigonometric function in the simplified form.

Complete step-by-step answer:

We have to find the maximum and minimum value of ${\cos ^6}\theta + {\sin ^6}\theta $.

So let’s simplify it first,

Let $f(\theta ) = {\cos ^6}\theta + {\sin ^6}\theta $.

So we can also write this as

$f(\theta ) = {\left( {si{n^2}\theta } \right)^3} + {\left( {{{\cos }^2}\theta } \right)^3}$

Now using ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$, we have

$f(\theta ) = \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)$

Using ${\sin ^2}\theta + {\cos ^2}\theta = 1$…………………………… (1)

$f(\theta ) = \left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)$

Now we can write $\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right) = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 3{\sin ^2}\theta {\cos ^2}\theta $

So we can write $f(\theta ) = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 3{\sin ^2}\theta {\cos ^2}\theta $

Now using equation 1 we have

$f(\theta ) = 1 - 3{\sin ^2}\theta {\cos ^2}\theta $

We can write this as

$f(\theta ) = 1 - \dfrac{3}{4} \times 4{\sin ^2}\theta {\cos ^2}\theta $

Now using $2\sin \theta \cos \theta = \sin 2\theta $

We have $f(\theta ) = \dfrac{3}{4}{\left( {\sin 2\theta } \right)^2}$

Using half angle formulae, $\left( {1 - \cos 4\theta } \right) = 2{\sin ^2}2\theta $

$ \Rightarrow 1 - \dfrac{3}{8}\left( {1 - \cos 4\theta } \right)$

Let’s simplify it further we get $1 - \dfrac{3}{8} + \dfrac{3}{8}\cos 4\theta $

Hence $f(\theta ) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\theta $……………………………. (2)

Now we know that $ - 1 \leqslant {\text{ cos4}}\theta {\text{ }} \leqslant {\text{ 1}}$ (maximum and minimum inbound of cos x)

$ \Rightarrow - \dfrac{3}{8} \leqslant {\text{ }}\dfrac{3}{8}{\text{cos4}}\theta {\text{ }} \leqslant {\text{ }}\dfrac{3}{8}$

$ \Rightarrow \dfrac{5}{8} - \dfrac{3}{8} \leqslant {\text{ }}\dfrac{5}{8} + \dfrac{3}{8}{\text{cos4}}\theta {\text{ }} \leqslant \dfrac{5}{8} + \dfrac{3}{8}$ (Adding $\dfrac{5}{{8{\text{ }}}}$to all sides of inequality)

Now using equation 2 we know that $f(\theta ) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\theta $

Hence

$\dfrac{1}{4} \leqslant f(\theta ) \leqslant 1$

Thus the minimum value of the required quantity is $\dfrac{1}{4}$and maximum value is 1

So option (a) is correct.

Note: Whenever we have to solve such problems, try to simplify as much as possible in order to reach the simplest form of expression, then use the min and max inbounds of the simplified part to reach up to the solution.

Complete step-by-step answer:

We have to find the maximum and minimum value of ${\cos ^6}\theta + {\sin ^6}\theta $.

So let’s simplify it first,

Let $f(\theta ) = {\cos ^6}\theta + {\sin ^6}\theta $.

So we can also write this as

$f(\theta ) = {\left( {si{n^2}\theta } \right)^3} + {\left( {{{\cos }^2}\theta } \right)^3}$

Now using ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$, we have

$f(\theta ) = \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)$

Using ${\sin ^2}\theta + {\cos ^2}\theta = 1$…………………………… (1)

$f(\theta ) = \left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)$

Now we can write $\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right) = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 3{\sin ^2}\theta {\cos ^2}\theta $

So we can write $f(\theta ) = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 3{\sin ^2}\theta {\cos ^2}\theta $

Now using equation 1 we have

$f(\theta ) = 1 - 3{\sin ^2}\theta {\cos ^2}\theta $

We can write this as

$f(\theta ) = 1 - \dfrac{3}{4} \times 4{\sin ^2}\theta {\cos ^2}\theta $

Now using $2\sin \theta \cos \theta = \sin 2\theta $

We have $f(\theta ) = \dfrac{3}{4}{\left( {\sin 2\theta } \right)^2}$

Using half angle formulae, $\left( {1 - \cos 4\theta } \right) = 2{\sin ^2}2\theta $

$ \Rightarrow 1 - \dfrac{3}{8}\left( {1 - \cos 4\theta } \right)$

Let’s simplify it further we get $1 - \dfrac{3}{8} + \dfrac{3}{8}\cos 4\theta $

Hence $f(\theta ) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\theta $……………………………. (2)

Now we know that $ - 1 \leqslant {\text{ cos4}}\theta {\text{ }} \leqslant {\text{ 1}}$ (maximum and minimum inbound of cos x)

$ \Rightarrow - \dfrac{3}{8} \leqslant {\text{ }}\dfrac{3}{8}{\text{cos4}}\theta {\text{ }} \leqslant {\text{ }}\dfrac{3}{8}$

$ \Rightarrow \dfrac{5}{8} - \dfrac{3}{8} \leqslant {\text{ }}\dfrac{5}{8} + \dfrac{3}{8}{\text{cos4}}\theta {\text{ }} \leqslant \dfrac{5}{8} + \dfrac{3}{8}$ (Adding $\dfrac{5}{{8{\text{ }}}}$to all sides of inequality)

Now using equation 2 we know that $f(\theta ) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\theta $

Hence

$\dfrac{1}{4} \leqslant f(\theta ) \leqslant 1$

Thus the minimum value of the required quantity is $\dfrac{1}{4}$and maximum value is 1

So option (a) is correct.

Note: Whenever we have to solve such problems, try to simplify as much as possible in order to reach the simplest form of expression, then use the min and max inbounds of the simplified part to reach up to the solution.

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