Question

The equation $y=\sin \left( x \right)\sin \left( x+1 \right)-{{\sin }^{2}}\left( x+1 \right)$ represents a straight line lying in ,$(A) second and third quadrants only$
(B) third and fourth quadrants only$(C) first, third and fourth quadrants$
(D) first, second and fourth quadrants$$$$

Answer 1

Use the product of two sine of angles formula $\sin A\sin B=\dfrac{1}{2}\left[ \cos \left( A-B \right)-\cos \left( A+B \right) \right]$
to simplify the first term $\sin \left( x \right)\sin \left( x+1 \right)$ and the formula $\dfrac{1-\cos \left( 2\theta \right)}{2}={{\sin }^{2}}\theta$ to simplify the second term ${{\sin }^{2}}\left( x+1 \right)$ in the right hand side of the given equation and proceed to simplify. You will obtain a linear equation $y=-{{\sin }^{2}}1$ after simplification. Use the sign of quadrants (for example in the first quadrant both abscissa and ordinate are positive) to find out the answer from the simplified equation.

Complete step-by-step answer:
The given trigonometric equation in sine is
$y=\sin \left( x \right)\sin \left( x+1 \right)-{{\sin }^{2}}\left( x+1 \right)...(1)$
We know from the trigonometric identity of cosine sum of two angles that,
$\sin A\sin B=\dfrac{1}{2}\left[ \cos \left( A-B \right)-\cos \left( A+B \right) \right]$
We also know the trigonometric identity of cosine twice of an angle,

& \cos \left( 2\theta \right)=1-2{{\sin }^{2}}\theta \\\ & \Rightarrow \dfrac{1-\cos \left( 2\theta \right)}{2}={{\sin }^{2}}\theta \\\ \end{aligned}$$ We take $x=B$,$x+2=A$ then $A-B=2$, $A+B=2x+2$ . Again if we take $x+1=\theta $ then $2\theta =2\left( x+1 \right)$ . We put these values in equation(1) and the equation transforms to $$\begin{aligned} & y=\sin \left( x \right)\sin \left( x+1 \right)-{{\sin }^{2}}\left( x+1 \right) \\\ & \Rightarrow y=\dfrac{1}{2}\left( \cos \left( 2 \right)-\cos \left( 2x+2 \right) \right)-2\left[ \dfrac{1-\cos \left( 2\left( x+1 \right) \right)}{2} \right] \\\ & \Rightarrow y=\dfrac{1}{2}\left( \cos \left( 2 \right)-1 \right) \\\ \end{aligned}$$ Again using the formula $\dfrac{1-\cos \left( 2\theta \right)}{2}={{\sin }^{2}}\theta $ in above we proceed to get, $$\begin{aligned} & \Rightarrow y=\dfrac{1}{2}\left( \cos \left( 2 \right)-1 \right) \\\ & \Rightarrow y=-2{{\sin }^{2}}1 \\\ \end{aligned}$$ The final equation obtained above is an equation of constant function. We know that any equation of the form $y=b$ where $b$ is real constant, is represented by a line parallel to $x$-axis in $xy$-plane. $$$$ So if we plot the obtained equation $y=-{{\sin }^{2}}1$ we will find that the equation represents the graph of a line parallel to $x$-axis . We know that the range of ${{\sin }^{2}}x$ is the set of positive real numbers so the value of $-{{\sin }^{2}}1$ will be negative. We know that the range of ${{\sin }^{2}}x$ is the set of positive real numbers so the value of $-{{\sin }^{2}}1$ will be negative. $$$$ We know that only in the third and fourth quadrants the value of $y$ is negative. So the line represented by $y=-{{\sin }^{2}}1$ will be a line parallel to $x$-axis and under $x$-axis in the third and fourth quadrants.

So, the correct answer is “Option B”.

Note: We need to be careful not to confuse the sign of the formulas. Here we have used $\sin A\sin B=\dfrac{1}{2}\left[ \cos \left( A-B \right)-\cos \left( A+B \right) \right]$ which can also be written as $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$ where $C=A+B,D=A-B$ Here we write reverse of the order first C then D unlike other formula $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right).$