Answer

Verified

450k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE