Question

For what and only what values of $\alpha$ lying between $0$ and $\pi$ is the inequality $\sin \alpha {\cos ^3}\alpha > {\sin ^3}\alpha \cos \alpha$valid?
A.$\alpha \in \left( {0,\dfrac{\pi }{4}} \right)$
B.$\alpha \in \left( {0,\dfrac{\pi }{2}} \right)$
C.$\alpha \in \left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)$
D.None of these

Answer 1

We will take all the trigonometric identities in the inequality to one side and solve it by using standard trigonometric identities. Since, the interval of $\alpha$ is given, we can use that and find the actual interval $\alpha$ for which the inequality $\sin \alpha {\cos ^3}\alpha > {\sin ^3}\alpha \cos \alpha$ is valid.

Complete answer:
The given inequality is
$\sin \alpha {\cos ^3}\alpha > {\sin ^3}\alpha \cos \alpha$.
Taking the ${\sin ^3}\alpha \cos \alpha$ to the left side of the inequality,
$\Rightarrow$ $\sin \alpha {\cos ^3}\alpha - {\sin ^3}\alpha \cos \alpha > 0$
Taking $\sin \alpha$ and $\cos \alpha$ common from the left side of the inequality,
$\Rightarrow$ $\sin \alpha \cos \alpha ({\cos ^2}\alpha - {\sin ^2}\alpha ) > 0$
We know that, ${\cos ^2}\alpha - {\sin ^2}\alpha = \cos 2\alpha$,
Substituting the value of ${\cos ^2}\alpha - {\sin ^2}\alpha = \cos 2\alpha$,
$\Rightarrow$ $\Rightarrow$ $\sin \alpha \cos \alpha (\cos 2\alpha ) > 0$
Multiplying and dividing the left side of the inequality by $2$,
$\Rightarrow$ $\dfrac{1}{2} \times 2 \times \sin \alpha \cos \alpha (\cos 2\alpha ) > 0$
We know that, $2\sin \alpha \cos \alpha = \sin 2\alpha$
Substituting the value of $2\sin \alpha \cos \alpha = \sin 2\alpha$,
$\Rightarrow$ $\dfrac{1}{2} \times \sin 2\alpha .\cos 2\alpha > 0$
Again, Multiplying and dividing the left side of the inequality by $2$,
$\Rightarrow$ $\dfrac{1}{2} \times \dfrac{1}{2} \times 2 \times \sin 2\alpha .\cos 2\alpha > 0$
$\Rightarrow$ $\dfrac{1}{4} \times 2 \times \sin 2\alpha .\cos 2\alpha > 0$
We know that, $2\sin 2\alpha \cos 2\alpha = \sin 4\alpha$
Substituting the value of $2\sin 2\alpha \cos 2\alpha = \sin 4\alpha$,
$\Rightarrow$ $\dfrac{1}{4} \times \sin 4\alpha > 0$
Taking $4$ to the right side of the inequality,
$\Rightarrow$ $\sin 4\alpha > 0 \times 4$
$\Rightarrow$ $\sin 4\alpha > 0$
Since, $0 < \alpha < \pi$ …. (Given)
We can say that,
$\Rightarrow$ $4\alpha \in (0,\pi )$
Dividing through by $4$ ,
$\Rightarrow$ $\alpha \in \left( {\dfrac{0}{4},\dfrac{\pi }{4}} \right)$
$\Rightarrow$ $\alpha \in \left( {0,\dfrac{\pi }{4}} \right)$
For value of $\alpha \in \left( {0,\dfrac{\pi }{4}} \right)$ the inequality $\sin \alpha {\cos ^3}\alpha > {\sin ^3}\alpha \cos \alpha$valid.
Therefore, the correct option is option A. $\alpha \in \left( {0,\dfrac{\pi }{4}} \right)$.

Note:
An inequality involving trigonometric functions of an unknown angle is called trigonometric inequality. A trig inequality is an inequality in preferred form: R(x) > 0 (or < 0) that consists of one or some trigonometric functions of the variable arc x. Fixing the inequality R(x) means finding all of the values of the variable arc x whose trig functions make the inequality R(x) real. This kind of value of x represents the answer set of the trig inequality R(x). Answer sets of trig inequalities are expressed in durations.