1. Given equation

We need to simplify $y=\sin x\sin(x+2)-\sin^2(x+1).$

2. Use product-to-sum identity

Recall, $\sin A\sin B=\frac{1}{2}\big[\cos(A-B)-\cos(A+B)\big].$

So, $\sin x\sin(x+2)=\frac{1}{2}\big[\cos(x-(x+2)) - \cos(x+(x+2))\big].$

That is, $\sin x\sin(x+2)=\frac{1}{2}\big[\cos(-2)-\cos(2x+2)\big].$

Since $\cos(-2)=\cos 2$, $\sin x\sin(x+2)=\frac{1}{2}\big[\cos 2-\cos(2x+2)\big].$

3. Use identity for $\sin^2\theta$

Recall, $\sin^2\theta=\frac{1-\cos 2\theta}{2}.$

Hence, $\sin^2(x+1)=\frac{1-\cos(2x+2)}{2}.$

4. Substitute into the expression

Now, $y=\frac{1}{2}\big[\cos 2-\cos(2x+2)\big]-\frac{1-\cos(2x+2)}{2}.$

Expand: $y=\frac{1}{2}\left[\cos 2-\cos(2x+2)-1+\cos(2x+2)\right].$

The $\cos(2x+2)$ terms cancel: $y=\frac{1}{2}(\cos 2-1).$

Thus $y$ is a constant, independent of $x$.

5. Interpret geometrically

So the graph is the horizontal line $y=\frac{\cos 2-1}{2}.$

Now, $\cos 2<1 \implies \cos 2-1<0,$ so $y<0.$

Also, $\cos 2\approx -0.416,$ therefore $y\approx \frac{-0.416-1}{2}=-0.708<0.$

Hence the line is a horizontal line below the $x$-axis.

6. Quadrants through which the line passes

A horizontal line with $y<0$ lies below the $x$-axis:

• for $x<0$, points are in the third quadrant,
• for $x>0$, points are in the fourth quadrant.

So the line lies in third and fourth quadrants only.

7. Check options

• A: first, second and fourth quadrants ❌
• B: first, third and fourth quadrants ❌
• C: second and third quadrants only ❌
• D: third and fourth quadrants only ✅

Therefore, the correct option is $\boxed{\text{D}}.$