# Find the maximum and minimum values of ${\cos ^6}\theta + {\sin ^6}\theta $ respectively.

$\left( a \right){\text{ 1 and }}\dfrac{1}{4}$

$\left( b \right){\text{ 1 and 0}}$

$\left( c \right){\text{ 2 and 0}}$

$(d){\text{ 1 and }}\dfrac{1}{2}$

Answer

Verified

381.9k+ 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

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

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

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

What is pollution? How many types of pollution? Define it

Give 10 examples for herbs , shrubs , climbers , creepers

Which planet is known as the red planet aMercury bMars class 6 social science CBSE

Which state has the longest coastline in India A Tamil class 10 social science CBSE

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