2^sin x + 2^cos x......where ^ is symbol for 'to the power'
a..certainly I can I can use differential and..but like if I take the 2 numbers in A.P.....then by using A.P.>GP...
$\displaystyle 2^{sin x}+2^{cos x} >=2^{1-1/sqrt(2)}$
so I am able to get the minimum value...but is there an algebraic way to solve it...not using calculus...I am rather uncomfortable with calculus if it goes complex.
sin(x) and cos(x) are equal when x = π/4. The value is $\displaystyle \frac{1}{\sqrt{2}$
sin(x) and cos(x) are equal but having opposite sign when x = 3π/4.
Hence maximum value of $\displaystyle 2^{sin(x)} + 2^{cos(x)} = 2\times2^\frac{1}{\sqrt{2}}$
Minimum value is $\displaystyle 2^\frac{1}{\sqrt{2}} + 2^\frac{-1}{\sqrt{2}}$