Question:
The equation \(y=\sin x\,\sin(x+2)-\sin^{2}(x+1)\) represents a straight line lying in :
The equation \(y=\sin x\,\sin(x+2)-\sin^{2}(x+1)\) represents a straight line lying in :
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Use \(\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]\) and \(\sin^2\theta=\frac{1-\cos2\theta}{2}\); the x-dependence cancels.
Updated On: Nov 5, 2026
- first, second and fourth quadrants
- first, third and fourth quadrants
- second and third quadrants only
- third and fourth quadrants only
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The Correct Option is D
Solution and Explanation
\(\sin x\sin(x+2)=\frac12[\cos2-\cos(2x+2)]\).
\(\sin^2(x+1)=\frac12[1-\cos(2x+2)]\).
Subtracting: \(y=\frac12[\cos2-1]=-\sin^2 1\).
This is a constant, so the graph is the horizontal line \(y=-\sin^21\), which is negative (and non-zero).
A horizontal line below the x-axis passes through the third and fourth quadrants only.
Answer: (4) third and fourth quadrants only
\(\sin^2(x+1)=\frac12[1-\cos(2x+2)]\).
Subtracting: \(y=\frac12[\cos2-1]=-\sin^2 1\).
This is a constant, so the graph is the horizontal line \(y=-\sin^21\), which is negative (and non-zero).
A horizontal line below the x-axis passes through the third and fourth quadrants only.
Answer: (4) third and fourth quadrants only
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