Question

Given expression: ${\sin ^6}x + {\cos ^6}x$ lies between
A) $\dfrac{1}{4}$and 1
B) $\dfrac{1}{4}$and 2
C) $0$ and 1
D) None of these

Answer 1

Hint: According to given in the question we have to solve the given expression ${\sin ^6}x + {\cos ^6}x$ so, first of all we have to convert the each term ${\sin ^6}x$ and ${\cos ^6}x$ in the form of cube.
Now, to solve the obtained trigonometric expression we have to use the formula as given below:

Formula used:
${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)..................(1)$
Hence with the help of the formula (1) we can simplify the obtained trigonometric expression but after that we have to multiply and divide with 4 with the obtained term of the expression to make it in the form of the formula as given below:
$2\sin A\sin B = \sin 2A................(2)$
Now, we can find the maximum and minimum values of the obtained trigonometric expression by substituting the maximum and minimum values of sin.

Complete step by step answer:
Step 1: First of all we have to convert each term ${\sin ^6}x$ and ${\cos ^6}x$ in the form of a cube.
$ = {({\sin ^2}x)^3} + {({\cos ^2}x)^3}$
Step 2: Now, we have to use the formula (1) as mentioned in the solution hint to determine or solve the expression as obtained in the step 1.
$ = {({\sin ^2}x + {\cos ^2}x)^3} - 3{\sin ^2}x{\cos ^2}x({\sin ^2}x + {\cos ^2}x)$
Step 3: On solving the expression as obtained in the step 2 and as we know that
${\sin ^2}x + {\cos ^2}x = 1$
Hence,
$
   = {(1)^3} - 3{\sin ^2}x{\cos ^2}x \times 1 \\
   = 1 - 3{\sin ^2}x{\cos ^2}x \\
 $
Step 4: Now, we have to multiply and divide by 4 in the term $3{\sin ^2}x{\cos ^2}x$ of the expression to use the formula (2) as mentioned in the solution hint.
$ = 1 - 3 \times \dfrac{4}{4}{\sin ^2}x{\cos ^2}x$
Now, we have to convert it into the form of formula (2),
$
   = 1 - \dfrac{3}{4} \times {(2\sin x\cos x)^2} \\
   = 1 - \dfrac{3}{4}{(\sin 2x)^2} \\
 $
Step 5: Now, we have to find the maximum and minimum value of the expression as obtained in step 4.
Maximum value of ${\sin ^6}x + {\cos ^6}x$ is $1 - \dfrac{3}{4} \times 0 = 1$
Minimum value of ${\sin ^6}x + {\cos ^6}x$ is $1 - \dfrac{3}{4} \times 1 = \dfrac{1}{4}$

Hence, with the help of the formula (1), and (2) we have obtained the value of the given trigonometric expression ${\sin ^6}x + {\cos ^6}x$ which lies between $\dfrac{1}{4}$ and 1. Therefore option (A) is correct.

Note:
It is necessary to convert the each term ${\sin ^6}x$ and ${\cos ^6}x$ in the form of the given expression in the form of cube as ${({\sin ^2}x)^3}$ and ${({\cos ^2}x)^3}$ so that we can easily simplified the given trigonometric expression.
To find the maximum and minimum value of the given trigonometric expression, we have to place the maximum and minimum values of sin lies between (0 to 1).