Find the maximum and minimum values of sin^4x + cos^2x and hence or otherwise find the maximum value of sin^1000 x + cox^2008 x

sin⁴x + cos²x  = (1-cos²x )² + cos²x 
= 1 + cos⁴x -2cos²x + cos²x
= 1+ cos⁴x -cos²x
= 1/4  + 3/4  + cos⁴x - cos²x
$$\begin{gathered} =({\cos }^{2}x -\frac{1}{2}{)}^2 +\frac{3}{4} \\ =(\frac{2{\cos }^{2}x -1}{2}{)}^2 +\frac{3}{4} \\ =\frac{(\cos 2x{)}^2}{4}+\frac{3}{4} \\ As (\cos 2x{)}^2 will lies between [0,1] \\ So above expression has minimum value when (\cos 2x{)}^2 = 0 \\ And maximum value when (\cos 2x{)}^2 = 1 \\ Hence the range is [\frac{3}{4},1] \end{gathered}$$

And sin¹⁰⁰⁰x  + cos²⁰⁰⁸x

Its maximum value will be 1 , as both sinx and cosx range are less than equal to one , so if a number less than 1 has exponents very high like 1000 or 2008 , then it will eventually lead to very small value and ultimately zero.
So the smallest value will be zero and maximum value will  be 1 , when x  = 0 or 90 degree .