Question

In which of the following sets the inequality $\sin^{6}x + \cos^{6}x > \dfrac{5}{8}$ holds good ?
A. $\left( - \dfrac{\pi}{8},\dfrac{\pi}{8} \right)$
B. $\left( \dfrac{3\pi}{8},\dfrac{5\pi}{8} \right)$
C. $\left( \dfrac{\pi}{4},\dfrac{3\pi}{4} \right)$
D. $\left( \dfrac{7\pi}{8},\dfrac{9\pi}{8} \right)$

Answer 1

In this question, we need to find which set holds good for the given inequality. First let us rewrite $\sin^{6}x + \cos^{6}x$ in the form of $a^{3} + b^{3}$ , then we can apply the formula of $a^{3} + b^{3} = \left( a + b \right)^{3} – 3ab\left( a + b \right)$ . Then we can use the trigonometric identity $\sin^{2}x + \cos^{2}x = 1$ , in order to simplify the expression. Then we can use a double angle identity that is $2\sin\ x\ \cos\ x\ = \sin\ 2x$ and $1 – 2\sin^{2}x = \cos\ 2x$ to solve the expression further. Finally we can try $n = 0,1,2$ . Using this we can find the interval of the given inequality.

Complete step-by-step answer:
Given, $\sin^{6}x + \cos^{6}x > \dfrac{5}{8}$
First let us rewrite the given expression as
$\Rightarrow \ \left( \sin^{2}x \right)^{3} + \left( \cos^{2}x \right)^{3} > \dfrac{5}{8}$
Since we know that $\left( a \right)^{6}$ can be written as $\left( a^{2} \right)^{3}$ .
Then we can apply $\left( a^{3} \right) + \left( b^{3} \right)$ formula.
That is
$\left( a^{3} \right) + \left( b^{3} \right) = \left( a + b \right)^{3} – 3ab\left( a + b \right)$
Thus we get,
$\Rightarrow \ \left( \sin^{2}x + cos^{2}x \right)^{3} – 3sin^{2}\ x\ \cos^{2}x\left( \sin^{2}x + \cos^{2}x \right) > \dfrac{5}{8}$
Now we know that $\sin^{2}x + \cos^{2}x = 1$
$\Rightarrow \ \left( 1 \right)^{3} – 3\sin^{2}\ x\ \cos^{2}x \times 1 > \dfrac{5}{8}$
On subtracting both sides by $1$ ,
We get,
$- 3\sin^{2}\ x\ \cos^{2}x > \dfrac{5}{8} – 1$
On taking LCM,
We get,
$- 3\sin^{2}\ x\ \cos^{2}x > \dfrac{5 – 8}{8}$
On simplifying,
We get,
$- 3\sin^{2}\ x\ \cos^{2}x > - \dfrac{3}{8}$
On dividing both sides by $- 3$ ,
We get,
$\sin^{2}\ x\ \cos^{2}x < \dfrac{1}{8}$
On cross multiplying,
We get,
$8\sin^{2}\ x \cos^{2}x < 1$
We also know that $2\sin\ x\ \cos\ x\ = \sin\ 2x$
$\Rightarrow \ 2\ \sin^{2}x < 1$
On subtracting both sides by $1$ ,
We get,
$\Rightarrow \ - 1 + 2\ \sin^{2}x < 0$
On dividing both sides by $( - )$ ,
We get,
$1 – 2\ \sin^{2}x > 0$
We know that $1 – 2\sin^{2}x = \cos\ 2x$
On applying this formula ,
We get,
$\Rightarrow \ \cos\ 4x > 0$
Where
$4x \in \left( 2n\pi - \dfrac{\pi}{2},\ 2n\pi + \dfrac{\pi}{2} \right)$
On simplifying,
We get,
$4x \in \left( \dfrac{4n\pi - \pi}{2},\dfrac{4n\pi + \pi}{2} \right)$
On dividing by $4$ ,
We get,
$\Rightarrow \ x \in \left( \dfrac{\left( \dfrac{4n\pi - \pi}{2} \right)}{4},\dfrac{\left( \dfrac{4n\pi + \pi}{2} \right)}{4} \right)$
On simplifying,
We get,
$x \in \left( \dfrac{4n\pi - \pi}{8},\dfrac{4n\pi + \pi}{8} \right)$
On taking $\pi$ common,
We get,
$x \in \left( \dfrac{\pi}{8}\left( 4n – 1 \right),\dfrac{\pi}{8}\left( 4n + 1 \right) \right)$
Now on substituting $n = 0$ ,
$\Rightarrow \ x \in \left( \dfrac{\pi}{8}\left( 4\left( 0 \right) – 1 \right),\dfrac{\pi}{8}\left( 4\left( 0 \right) + 1 \right) \right)\$
On simplifying,
We get,
$\Rightarrow \ x \in \ \left( \dfrac{\pi}{8}, - \dfrac{\pi}{8} \right)$
Now we can try $n = 1$ ,
On substituting $n = 1$ ,
We get
$\Rightarrow \ x \in \ \left( \dfrac{\pi}{8}\left( 4\left( 1 \right) – 1 \right),\dfrac{\pi}{8}\left( 4\left( 1 \right) + 1 \right) \right)$
On simplifying,
We get,
$x \in \left( \dfrac{\pi}{8}\left( 3 \right),\dfrac{\pi}{8}\left( 5 \right) \right)$
On further simplifying,
We get,
$x \in \left( \dfrac{3\pi}{8},\dfrac{5\pi}{8} \right)$
Now let us substitute $n = 2$ ,
$\Rightarrow \ x \in \ \left( \dfrac{\pi}{8}\left( 4\left( 2 \right) – 1 \right),\dfrac{\pi}{8}\left( 4\left( 2 \right) + 1 \right) \right)$
On simplifying,
We get,
$x \in \left( \dfrac{\pi}{8}\left( 7 \right),\dfrac{\pi}{8}\left( 9 \right) \right)$
On further simplifying,
We get,
$x \in \left( \dfrac{7\pi}{8},\dfrac{7\pi}{8} \right)$
Thus we get the inequality $\sin^{6}x + \cos^{6}x > \dfrac{5}{8}$ holds good in $\left( \dfrac{\pi}{8}, - \dfrac{\pi}{8} \right)$ , $\left( \dfrac{3\pi}{8},\dfrac{5\pi}{8} \right)$ and $\left( \dfrac{7\pi}{8},\dfrac{7\pi}{8} \right)$
Final answer :
The inequality $\sin^{6}x + \cos^{6}x > \dfrac{5}{8}$ holds good in $\left( \dfrac{\pi}{8}, - \dfrac{\pi}{8} \right)$ , $\left( \dfrac{3\pi}{8},\dfrac{5\pi}{8} \right)$ and $\left( \dfrac{7\pi}{8},\dfrac{7\pi}{8} \right)$

Option A, B and D is the correct answer.

Note: In order to solve these types of questions, we must have a stronger grip over the trigonometric identity and properties . We should also be careful while using the double angle identity in order to solve the expression. We need to note that when we are multiplying or dividing by $( - )$ with the inequality , then the direction of inequality gets reversed because when any negative number is multiplied to the inequality, its direction changes. We should also be careful in converting the term $\sin^{6}x$ and $\cos^{6}x$ in the form $\left( a^{2} \right)^{3}$ as $\left( \sin^{2}x \right)^{3}$ and $\left( \cos^{2}x \right)^{3}$ so that we can easily simplified the given trigonometric expression.