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

Hint: 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.