The equation $$y=\sin x\sin\left(x+2\right)-\sin^2(x+1)$$ represents a straight line lying in:

We start from the given relation

$$y=\sin x\,\sin (x+2)\;-\;\sin^{2}(x+1).$$

First we simplify the product $$\sin x\,\sin (x+2)$$ by recalling the product-to-sum identity

$$\sin A\,\sin B=\frac{\cos(A-B)-\cos(A+B)}{2}.$$

Taking $$A=x,\;B=x+2,$$ we get

$$\sin x\,\sin (x+2)=\frac{\cos\!\bigl(x-(x+2)\bigr)-\cos\!\bigl(x+(x+2)\bigr)}{2} =\frac{\cos(-2)-\cos(2x+2)}{2} =\frac{\cos 2-\cos(2x+2)}{2}.$$

Substituting this result back, the expression for y becomes

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

Next we simplify $$\sin^{2}(x+1)$$ with the double-angle formula

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

Putting $$\theta=x+1,$$ we have

$$\sin^{2}(x+1)=\frac{1-\cos\!\bigl(2x+2\bigr)}{2}.$$

Inserting this into the previous expression gives

$$ \begin{aligned} y &= \frac{\cos 2-\cos(2x+2)}{2}\;-\;\frac{1-\cos(2x+2)}{2} \\[4pt] &=\frac{\cos 2-\cos(2x+2)-1+\cos(2x+2)}{2} \\[4pt] &=\frac{\cos 2-1}{2}. \end{aligned} $$

Observe that every term containing x has cancelled out; hence y is a constant. To see its sign, rewrite

$$\cos 2-1=-(1-\cos 2).$$

Using $$1-\cos\theta=2\sin^{2}\dfrac{\theta}{2},$$ with $$\theta=2,$$ we get

$$1-\cos 2=2\sin^{2}1,$$

so

$$\cos 2-1=-2\sin^{2}1.$$

Therefore

$$y=\frac{\cos 2-1}{2}= \frac{-2\sin^{2}1}{2}=-\sin^{2}1.$$

The quantity $$\sin^{2}1$$ is positive (because $$1\text{ rad}\approx57^\circ$$ lies in the first quadrant). Hence

$$y=-\sin^{2}1\lt 0\quad\text{for every }x\in\mathbb R.$$

Thus every point of the straight line has a negative ordinate. Since x can take both positive and negative values, points on the line occur with

• $$x \lt 0,\;y \lt 0$$  ⇒ third quadrant, and
• $$x \gt 0,\;y \lt 0$$  ⇒ fourth quadrant.

No point can appear in the first or second quadrants because there $$y\gt 0.$$

Hence, the correct answer is Option D.