Question

Range of ${{\sin }^{3}}x-{{\sin }^{3}}({{360}^{\circ }}-3x)+{{\sin }^{3}}({{360}^{\circ }}+3x)$
A.$\left[ -1,1 \right]$
B.$\left[ \dfrac{-1}{4},\dfrac{1}{4} \right]$
C.$\left[ \dfrac{-3}{4},\dfrac{3}{4} \right]$
D.None of these.

Answer 1

Hint : We will first simplify the given expression before finding the range since that will make things easier. We will change ${{\sin }^{3}}({{360}^{\circ }}-3x)$ using trigonometric ratios.

Complete step-by-step answer :
First, we should know what trigonometric ratios are.
Trigonometric ratio says that whenever there is an odd integer with 90 we will change the trigonometric function such as, sine will change to cosine and vice versa. Tan will change to sec and vice versa. Cot will change to cosec and vice versa. If the number is even then no change will take place.
$\sin (n\times {{90}^{\circ }}+\theta )$

where n is numeric whole integer and $\theta$ is the angle.
example is $\sin (2\times {{90}^{\circ }}-{{30}^{\circ }})=\sin ({{30}^{\circ }})$
example $\sin (3\times {{90}^{\circ }}-{{30}^{\circ }})=-\cos {{30}^{\circ }}$
If n is 1 then it means that it is in the 1st quadrant where all the value of any trigonometric function is positive. $n=2$ means in the second, $n=3$ means 3rd and $n=4$means 4th quadrant. After $n=5$ then it means that we have made 360 thus we will be once again in the 1st quadrant. And this goes on.
Here n=4 so we are initially in the 4th quadrant where cosine and sec values are only positive others are negative.
Next thing we have to observe is what the value of $\theta$ will do. if the value of $\theta$ is adding up with . $(4\times 90)$ then it means that we are going to 1st quadrant and if we see that the value of $\theta$is subtracted from $(4\times 90)$ then it means that we are still in the 4th quadrant.
Thus, after taking all of this into note we will see that:
${{\sin }^{3}}x-{{\sin }^{3}}({{360}^{\circ }}-3x)+{{\sin }^{3}}({{360}^{\circ }}+3x)$
${{\sin }^{3}}x-{{(\sin (4\times 90-3x))}^{3}}+{{(\sin (4\times 90+3x))}^{3}}$
${{\sin }^{3}}x-{{(-\sin (3x))}^{3}}+{{(\sin (3x))}^{3}}\text{ }\\!\\![\\!\\!\text{ }\sin (4\times 90-3x)=-\sin (3x)\text{ because this lies in the}$
$\text{ 4th quadrant where only cos and sec is positive}\text{.}$
$\text{ }\sin (4\times 90+3x)=\sin 3x\text{ because thies comes to the }$
$\text{ 1st quadrant where everything is positive}\text{. }\\!\\!]\\!\\!\text{ }$
$\text{si}{{\text{n}}^{3}}x+{{\sin }^{3}}3x+{{\sin }^{3}}3x$
${{\sin }^{3}}x+2({{\sin }^{3}}3x)-(1)$
Thus, after simplifying we are left with ${{\sin }^{3}}x+2({{\sin }^{3}}3x)$.
We can let $y={{\sin }^{3}}x+2({{\sin }^{3}}3x)-(1)$
We also know the formula of $\sin 3\theta$ which is$\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta$
${{\sin }^{3}}\theta =\dfrac{3\sin \theta }{4}-\dfrac{\sin 3\theta }{4}-(2)$
Thus using it on $y={{\sin }^{3}}x+2({{\sin }^{3}}3x)-(1)$
we get:

From (2), we can say that (1)=
$y={{\sin }^{3}}x+2({{\sin }^{3}}3x)$
$y=\dfrac{3\sin x}{4}-\dfrac{\sin 3x}{4}+2\left( \dfrac{3\sin (3x)}{4}-\dfrac{\sin 3(3x)}{4} \right)$
$y=\dfrac{3\sin x}{4}-\dfrac{\sin 3x}{4}+\dfrac{6\sin (3x)}{4}-\dfrac{2\sin 9x}{4}$
$y=\dfrac{3\sin x}{4}-\sin 3x\left( -\dfrac{1}{4}+\dfrac{6}{4} \right)-\dfrac{2\sin 9x}{4}$
$y=\dfrac{3\sin x}{4}+\dfrac{5\sin 3x}{4}-\dfrac{2\sin 9x}{4}$
Now, we have $y=\dfrac{3\sin x}{4}+\dfrac{5\sin 3x}{4}-\dfrac{2\sin 9x}{4}$
Now we can find the range of $y$ which is the limiting value of $y$ for it to be true or valid.
$\text{we know that }-1\langle \,\,\,\sin x\,\,\,\langle \,\,+1$
$-\dfrac{3}{4}\langle \,\,\,\dfrac{3}{4}\sin x\,\,\,\langle \,\,+\dfrac{3}{4}$
$-\dfrac{5}{4}\langle \,\,\,\dfrac{5}{4}\sin 3x\,\,\,\langle \,\,\dfrac{5}{4}$
$-\dfrac{2}{4}\langle \,\,\,\sin 9x\,\,\,\langle \,\,+\dfrac{2}{4}$
So the lowest parameter or the lowest value of y can be when all the three sin function have lowest value. Thus, when all are lowest:
$y=\dfrac{3\sin x}{4}+\dfrac{5\sin 3x}{4}-\dfrac{2\sin 9x}{4}$
$y=\dfrac{-3}{4}+\dfrac{-5}{4}-\left( \dfrac{-2}{4} \right)$
$y=\dfrac{-3-5+2}{4}=-\dfrac{6}{4}=-\dfrac{3}{2}$
when y is at maximum:
$y=\dfrac{3\sin x}{4}+\dfrac{5\sin 3x}{4}-\dfrac{2\sin 9x}{4}$
$y=\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{2}{4}=\dfrac{3+5-2}{4}=\dfrac{6}{4}=\dfrac{3}{2}$
$y=\left[ \dfrac{-3}{2},\dfrac{3}{2} \right]$
Thus the range is $y=\left[ \dfrac{-3}{2},\dfrac{3}{2} \right]$. This means that option D is correct. None of these.
So, the correct answer is “Option D”.

Note : The range of $y$ means its maximum and minimum value which limits it between these parameters. And this is also the true meaning of range. Meaning where the value of y can be true.