Question

Find the greatest and least value of (sin-1x)3+(cos​-1x)3?

Answer 1

INVERSE TRIGONOMETRIC FUNCTIONS & APPLICATION OF DERIVATIVES  :D 

Final Answer : Min. Value :  $\frac{\pi^{3}}{32}$  
                         Max . Value : $\frac{7\pi^{3}}{8}$ 

STEPS: 
1)   Let 
[tex]f(x) = (sin^{-1}x )^{3} \:\: + (cos^{-1}x)^{3} \\ \\ f'(x) = 3 (sin^{-1}x)^{2} * \frac{1}{ \sqrt{1-x^{2}} } + 3 (cos^{-1}x)^{2} * (- \frac{1}{ \sqrt{1-x^{2}} }) \\ \\ f'(x) = \frac{3}{ \sqrt{1-x^{2}} } ((sin^{-1}x)^{2}- (cos^{-1}x)^{2} ) \\ \\ [/tex]
 

Now ,Find below points as your own or take it as your Homework  which is trivial 
Critical Points / Boundary Points : $x= 1,-1, \frac{1}{ \sqrt{2} }$
 
 2)  Since we know ,
arcsin(1) = pi/2 , arcsin(1/root(2)) = pi/4 
arccos(-1) =pi   ,  arccos(1/root(2)) = pi/4 

Substitute these values in  f(x) , to get   
 f(-1) =   $\frac{7\pi^{3}}{8}$ 
    f(1)  =   $\frac{\pi^{3}}{8}$
   f(1/root(2))  = $\frac{ \pi^{3} }{32}$ 

3) Hence , looking above values :

Max f(x)  =  $\frac{7\pi^{3}}{8}$ 
Min f(x)    =  $\frac{ \pi^{3} }{32}$  

Domain  :  $-1 \leq x \leq 1$  

Hope its  Clear :p