Graph of the function $\cos(x)\cos(x+2)-\cos^2(x+1)$ will be?
(A)A straight line (B)A parabolaGive the corresponding equation too.
Source:JEE 1997.
Can someone suggest how should I proceed?Can't think of any way to reduce the the given equation to a straight line or a parabola's equation.Please guide me!
On
Let $f(x)=\cos(x)\cos(x+2)-\cos^2(x+1)$
\begin{align} f'(x)&=-\sin x\cos(x+2)-\cos x\sin(x+2)+2\cos(x+1)\sin(x+1)\\ &=-\sin(2x+2)+\sin(2x+2)\\ &=0 \end{align}
So it is a constant function, i.e., a horizontal straight line.
On
$$\cos(x)=\cos(x+1-1)=\cos(x+1)\cos(1)+\sin(x+1)\sin(1)$$ we use the fact $\cos(-1)=\cos(1)$, hence $\cos(x)$ is a even function, and we have too: $$\cos(x+2)=\cos(x+1+1)=\cos(x+1)\cos(1)-\sin(x+1)\sin(1)$$ then $$\cos(x)\cos(x+2)=\cos^{2}(x+1)\cos^{2}(1)-\sin^{2}(x+1)\sin^{2}(1)=$$ $$=\cos^{2}(x+1)\cos^{2}(1)-\sin^{2}(x+1)(1-\cos^{2}(1))=\cos^{2}(1)-\sin^{2}(x+1)$$ we now come back to the main formula $$\cos(x)\cos(x+2)-\cos^{2}(x+1)=\cos^{2}(1)-\sin^{2}(x+1)-\cos^{2}(x+1)$$ $$=\cos^{2}(1)-1=-\sin^{2}(1)$$ finally $$\cos(x)\cos(x+2)-\cos^{2}(x+1)=-\sin^{2}(1)$$
On
I think the easiest approach is to evaluate your expression at $x=0,x=-1,x=-2$ and recall that three collinear points cannot lie on a parabola. Even easier: your expression is bounded for sure between $-2$ and $1$. Given the options, it has to be a line, and a horizontal one: the same conclusion also follows from noticing that your function is $2\pi$-periodic.
\begin{align*} 2\cos(x)\cos(x+2) - 2\cos^2(x+1) &= \cos(2x+2)+\cos(2) - (1+\cos(2(x+1)) \\ &= \cos(2) - 1 \end{align*} Hence the graph is a straight line parallel to the $x$ axis.