Here's the question you clicked on:
laerb911
what is the tangent line of F(x)= sin^2 x at points (pi/4,1/2)
Find the derivative for your function to find the slope at pi/4
\[F(x)=\sin^2x\]\[F '(x)=2\sin x\cos x\]\[F'(\pi/4)=2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\]\[=2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)=1\]
now you have the slope (I think haha) and then you also have a point \((\pi/4,1/2)\) so then you have \[\pi/4+b=1/2\]\[b = 1/2-\pi/4\]so \(g(x)=x+1/2-\pi/4\) is the tangent line. At least I think. I have a bad hangover.