The equation y = sin ,x sin ,(x + 2) - sin^2 | vedclass

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(C) Given equation: $y = \sin \,x \sin \,(x + 2) - \sin^2 \,(x + 1)$Multiply by $2$: $2y = 2 \sin \,x \sin \,(x + 2) - 2 \sin^2 \,(x + 1)$Using the id

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third and fourth quadrants only

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(C) Given equation: $y = \sin \,x \sin \,(x + 2) - \sin^2 \,(x + 1)$Multiply by $2$: $2y = 2 \sin \,x \sin \,(x + 2) - 2 \sin^2 \,(x + 1)$Using the identity $2 \sin \,A \sin \,B = \cos(A - B) - \cos(A + B)$:$2 \sin \,x \sin \,(x + 2) = \cos(x - (x + 2)) - \cos(x + x + 2) = \cos(-2) - \cos(2x + 2) = \cos \,2 - \cos(2x + 2)$Using the identity $2 \sin^2 \,A = 1 - \cos(2A)$:$2 \sin^2 \,(x + 1) = 1 - \cos(2(x + 1)) = 1 - \cos(2x + 2)$Substituting these into the equation:$2y = (\cos \,2 - \cos(2x + 2)) - (1 - \cos(2x + 2))$$2y = \cos \,2 - 1$Since $\cos \,2 \approx -0.416$,we have $2y \approx -1.416$,so $y \approx -0.708$.This represents a horizontal line $y = k$ where $k $A$ horizontal line with a negative $y$-intercept passes through the third and fourth quadrants.

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

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