Question

The maximum value of ${{\sin }^{3}}x+{{\cos }^{3}}x$ is

Answer 1

Hint: To solve this question, firstly we will find the derivative of the function $f(x)={{\sin }^{3}}x+{{\cos }^{3}}x$and by putting the first derivative equal to zero, we will find extreme points. Then, we will find the second derivative of the function $f(x)={{\sin }^{3}}x+{{\cos }^{3}}x$, and for both extreme values we will check the value of the second derivative and accordingly we will verify which point will be the point of maxima. Then, we will evaluate the maximum value of function $f(x)={{\sin }^{3}}x+{{\cos }^{3}}x$, by putting that point of maxima in function.

Complete step-by-step solution
Now, let function be
$f(x)={{\sin }^{3}}x+{{\cos }^{3}}x$
We know that $\dfrac{d}{dx}{{f}^{n}}(x)=n{{f}^{n-1}}(x)f'(x)$
Differentiating, $f(x)={{\sin }^{3}}x+{{\cos }^{3}}x$ with respect to x, we get
$f'(x)=3{{\sin }^{2}}x\cos x-3{{\cos }^{2}}x\sin x$
As we know that, $\dfrac{d}{dx}(\sin ax)=a\cos ax$ and $\dfrac{d}{dx}(\cos ax)=-a \ sin ax$
Now, putting $f'(x)=3{{\sin }^{2}}x\cos x-3{{\cos }^{2}}x\sin x$ equals to zero to get extreme points, we get
$3{{\sin }^{2}}x\cos x-3{{\cos }^{2}}x\sin x=0$
On re-writing terms, we get
$3\cos x\sin x(\sin x-\cos x)=0$
$\cos x\sin x(\sin x-\cos x)=0$
Multiplying and dividing the term by 2, we get
$\dfrac{2}{2}\cos x\sin x(\sin x-\cos x)=0$
We know that 2sinxcosx = sin2x
So, we get
$\dfrac{\sin 2x(\sin x-\cos x)}{2}=0$
Or, $\sin 2x(\sin x-\cos x)=0$
So, we will have two cases, which are
sin2x = 0 or sinx – cosx = 0
$\Rightarrow$ sin2x = 0 or sinx = cosx
$\Rightarrow$ sin2x = 0 or tanx = 1
now, in interval of $\left( 0,\dfrac{\pi }{2} \right)$ , we have
sin2x = 0 for x = 0, and
tanx = 1 for $x=\dfrac{\pi }{4}$ .
so, now differentiating $f'(x)=3{{\sin }^{2}}x\cos x-3{{\cos }^{2}}x\sin x$, with respect to x, which will give second derivative of function f(x), we get
$f''(x)=\dfrac{d}{dx}\left( 3{{\sin }^{2}}x\cos x-3{{\cos }^{2}}x\sin x \right)$
We know that, $\dfrac{d}{dx}f(x).g(x)=f'(x)g(x)+g'(x)f(x)$
So, on solving we get
$f''(x)=3(2\sin x{{\cos }^{2}}x-{{\sin }^{3}}x)-3(-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x)$
$\Rightarrow f''(x)=6\sin x{{\cos }^{2}}x+6\cos x{{\sin }^{2}}x-3{{\cos }^{3}}x-3{{\sin }^{3}}x$
Now, at point x = 0,
$f''(0)=6\sin (0){{\cos }^{2}}(0)+6\cos (0){{\sin }^{2}}(0)-3{{\cos }^{3}}(0)-3{{\sin }^{3}}(0)$
$\Rightarrow f''(0)=-3<0$, as we know that sin(0) = 0 and cos(0) = 1
Now, at point $x=\dfrac{\pi }{4}$
$f''\left( \dfrac{\pi }{4} \right)=6\sin \left( \dfrac{\pi }{4} \right){{\cos }^{2}}\left( \dfrac{\pi }{4} \right)+6\cos \left( \dfrac{\pi }{4} \right){{\sin }^{2}}\left( \dfrac{\pi }{4} \right)-3{{\cos }^{3}}\left( \dfrac{\pi }{4} \right)-3{{\sin }^{3}}\left( \dfrac{\pi }{4} \right)$
\[\Rightarrow f''\left( \dfrac{\pi }{4} \right)=6\left( \dfrac{1}{\sqrt{2}} \right){{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+6\left( \dfrac{1}{\sqrt{2}} \right){{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-3{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}-3{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}\], as we know that
\[\Rightarrow \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] and \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]
On simplifying, we get
\[f''\left( \dfrac{\pi }{4} \right)=6\left( \dfrac{1}{\sqrt{2}} \right){{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+6\left( \dfrac{1}{\sqrt{2}} \right){{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-6{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}\]
\[\Rightarrow f''\left( \dfrac{\pi }{4} \right)=6\left( \dfrac{1}{\sqrt{2}} \right){{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\]
\[\Rightarrow f''\left( \dfrac{\pi }{4} \right)=\left( \dfrac{3}{\sqrt{2}} \right)>0\]
Now, we know that if f’’(x) < 0, at x = c then x = c is a point of maxima and if f’’(x) > 0, at x = c then x = c is a point of minima.
So, x = 0 is a point of maxima.
Then, at x = 0
$f(0)={{\sin }^{3}}(0)+{{\cos }^{3}}(0)$
f(0) = 0 + 1 = 1
Hence, the maximum value of ${{\sin }^{3}}x+{{\cos }^{3}}x$ is 1.

Note: To solve such a type of question, one of the best ways to solve these questions is a second derivative test. Always remember that $\dfrac{d}{dx}{{f}^{n}}(x)=n{{f}^{n-1}}(x)f'(x)$, $\dfrac{d}{dx}(\sin ax)=a\cos ax$, $\dfrac{d}{dx}(\cos ax)=-a \sin ax$, $\dfrac{d}{dx}f(x).g(x)=f'(x)g(x)+g'(x)f(x)$. Try not to make any calculation errors while solving the question.