Question

The equation y = sinx + sin(x + 2) - sin² (x +1) represents a straight line lying in

• A. first, third and fourth quadrants
• B. third and fourth quadrants only
• C. second and third quadrants only
• D. first, second and fourth quadrants

Answer 1

None of the options are correct as the equation does not represent a straight line.

Answer 2

The given equation is $y = \sin x + \sin(x + 2) - \sin^2 (x +1)$.

Step 1: Simplify the expression using trigonometric identities. First, apply the sum-to-product formula for $\sin A + \sin B$: $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$ Let $A = x$ and $B = x+2$. Then, $\frac{A+B}{2} = \frac{x + (x+2)}{2} = \frac{2x+2}{2} = x+1$. And, $\frac{A-B}{2} = \frac{x - (x+2)}{2} = \frac{-2}{2} = -1$. So, $\sin x + \sin(x+2) = 2 \sin(x+1) \cos(-1)$. Since $\cos(-\theta) = \cos(\theta)$, we have $\cos(-1) = \cos(1)$. Thus, $\sin x + \sin(x+2) = 2 \sin(x+1) \cos(1)$.

Substitute this back into the original equation for $y$: $y = 2 \sin(x+1) \cos(1) - \sin^2 (x +1)$.

Step 2: Analyze the nature of the simplified expression. The problem states that the equation represents a straight line. A straight line is generally represented by $y = mx+c$. If $m \neq 0$, $y$ varies linearly with $x$. If $m=0$, $y$ is a constant ($y=c$), representing a horizontal line. The expression $y = 2 \sin(x+1) \cos(1) - \sin^2(x+1)$ is a function of $\sin(x+1)$. Let $S = \sin(x+1)$. The equation becomes $y = 2S \cos(1) - S^2$. This is a quadratic function of $S$. Since $S = \sin(x+1)$ varies as $x$ varies (its range is $[-1, 1]$), $y$ will also vary. Therefore, $y$ is not a constant, and it is also not a linear function of $x$ (it is a periodic function).

Step 3: Determine the range of y to check for consistency with the options. The range of $\sin(x+1)$ is $[-1, 1]$. The function $f(S) = -S^2 + 2\cos(1)S$ is a downward-opening parabola. The vertex occurs at $S = -\frac{2\cos(1)}{2(-1)} = \cos(1)$. Since $1$ radian $\approx 57.3^\circ$, $\cos(1)$ is a positive value less than 1 (approximately $0.54$). This value lies within the range $[-1, 1]$. The maximum value of $y$ occurs at $S = \cos(1)$: $y_{max} = 2\cos(1) \cdot \cos(1) - \cos^2(1) = 2\cos^2(1) - \cos^2(1) = \cos^2(1)$. Numerically, $\cos^2(1) \approx (0.54)^2 \approx 0.29$.

The minimum value of $y$ occurs at one of the endpoints of the range of $S$, i.e., $S=-1$ or $S=1$. At $S=1$: $y = 2\cos(1) - 1 \approx 2(0.54) - 1 = 1.08 - 1 = 0.08$. At $S=-1$: $y = 2(-1)\cos(1) - (-1)^2 = -2\cos(1) - 1 \approx -2(0.54) - 1 = -1.08 - 1 = -2.08$. The minimum value of $y$ is approximately $-2.08$.

So, the range of $y$ is approximately $[-2.08, 0.29]$. This means $y$ can take positive values (e.g., $0.29, 0.08$), negative values (e.g., $-2.08$), and also $y=0$ (when $\sin(x+1)=0$, i.e., $x+1=n\pi$).

Step 4: Conclude based on the analysis and options. Since $y$ varies and is not a constant, the equation $y = \sin x + \sin(x + 2) - \sin^2 (x +1)$ does not represent a straight line. The premise of the question is false.

If the question implies that the graph of the function lies in certain quadrants:

• When $y>0$ (e.g., $y=0.1$), the graph lies in the first and second quadrants.
• When $y<0$ (e.g., $y=-1$), the graph lies in the third and fourth quadrants.
• When $y=0$, the graph lies on the x-axis (boundary of all four quadrants). Therefore, the graph of the function passes through all four quadrants. None of the given options (1), (2), (3), (4) represent "all four quadrants".

Given the contradiction between the problem statement ("represents a straight line") and the mathematical derivation (the expression is not a constant), the question is flawed. In such cases in competitive exams, if a choice must be made, it usually implies a constant value that was intended. However, based on the derivation, no such constant value is consistently represented by the expression for all $x$.

The final answer is $\boxed{None of the options are correct as the equation does not represent a straight line.}$