Answer
$y=1$
Work Step by Step
Let $\log x$ denote the natural logarithm $\ln x$.
$y=(sinx)^{2x}$
$y’=(sinx)^{2x}(\frac{d}{dx}(\log (sinx)2x)$
$u=\log(sinx)$
$u’=\frac{cos x}{sin x}$
$v=2x$
$v’=2$
$y’=(sinx)^{2x}(2x cotx +2 \log(sin x))$
$y’(\frac{\pi}{2})=2(sin(\frac{\pi}{2}))^{2\frac{\pi}{2}}(\frac{\pi}{2}cot(\frac{\pi}{2})+2\log(sin(\frac{\pi}{2}))$
$=0$
equation of tangent:
$y-1=0(x-\frac{\pi}{2})$
$y=1$