Trignometric functions, Maximum value $f(x)=\sin^{16}(x) +\cos^{18}(x)$,
find the maximum value of the given function $f(x)$.
I tried differentiating the given function however couldn't get the value of $x$ for which I could substitute the value in the given function so as to get the maximum value.
Please answer this query with detailed analysis 
thanking you in advance
sudhanshu singh
 A: Let $a = \sin^2x$, and $b = \cos^2x$, then $a + b = 1$, and $f(x) = f(a) = a^8 + b^9 = a^8 + (1-a)^9$, for $0 \leq a \leq 1$. You can find $f'(a)$, and set it equal to $0$ to continue.
A: As bounded functions, we have $ \ 0 \ \le \ \sin^{16} \ x \  \le \ 1 \ $ and $ \ 0 \ \le \ \cos^{18} \ x \ \le \ 1 \ $ .  The maximum for each function is located exactly at values of $ \ x \ $ where the other equals zero.  Moreover, the value of either function falls to less than $ \ \left( 0.9 \right)^{16} \ \approx \ 0.185 \ $  within 0.5 units of their individual maxima, which lie $ \ \frac{\pi}{2} \ \approx \ 1.57 \ $ units apart.  Midway between each maximum of $ \ f(x) \ $, either function is less than or equal to $ \ \left(\frac{1}{\sqrt{2}}\right)^{16} \ = \ \frac{1}{2^8} \ $ .  Each function strictly decreases to either side of its maxima in proceeding to their minima. $ ^* $ So the sum of the two never exceeds 1.  
$  * \ $ We will need to resort to calculus to show that the absolute value of the slope of either function is at least $ \ \frac{1}{2} \ $ over intervals from a few-hundredths to half-a-unit (or so) away from the maximum of each function.
