Answer
The identity is verified.
$sin^{\frac{1}{2}}x~cos~x-sin^{\frac{5}{2}}x~cos~x=cos^3x\sqrt {sin~x}$
Work Step by Step
$cos^2x+sin^2x=1$
$cos^2x=1-sin^2x$
$sin^{\frac{1}{2}}x~cos~x-sin^{\frac{5}{2}}x~cos~x=sin^{\frac{1}{2}}x~cos~x-sin^{\frac{1}{2}}x~sin^2x~cos~x=sin^{\frac{1}{2}}x~cos~x(1-sin^2x)=sin^{\frac{1}{2}}x~cos~x~cos^2x=sin^{\frac{1}{2}}x~cos^3x=cos^3x\sqrt {sin~x}$
