Can you find... trigonometric edition part 2

Well done to Olmo from Hockerill Anglo-European College for solving this problem. Here is his solution.

Trigonometric graphs take the form $ y = a \sin {(bx-c)} + d$, and similarly for cos and tan.

Can you find ...

(a) ... two sine graphs which only cross each other on the x-axis?

Any two graphs that simply have a different value for $a$ will fulfil these conditions, since the period and the intersections are the same, the only thing changed is the height of them.

e.g.

$y=\sin{x}$ and $y=-5\sin{x}$

or

$y=0.5\sin{x}$ and $y=3\sin{x}$

(b) ... a sine graph, a cosine graph and a tangent graph which all meet at certain points? What if all three graphs have to meet at the origin?

Manipulating the graphs by translating them gives many solutions.

e.g.

$y=\sin{x}$, $y=\cos{x}+1$, and $y=\tan{(x-\frac{\pi}{4})}$ intersect at the point $x=\frac{\pi}{2}$, $y=1$

or

$y=\sin{x}$, $y=\cos{(x-\frac{\pi}{2})}$, and $y=\tan{x}$ cross each other at the origin.

(c) ... a sine graph and a cosine graph which don't cross each other? What if the graphs have to lie between y=1 and y=−1?

In this case, the amplitude has to be changed (decreased), otherwise they will both reach 1 and -1 (or pass it) thus intersecting with each other. Then the graphs need adjusting, for example, the sine graph to the same shape as the cosine, but below it. So it will have a vertical and horizontal translation.

e.g.

$y=\frac{1}{2} \sin{(x+\frac{\pi}{2})}-\frac{1}{4}$ and $y=\frac{1}{2}\cos{x}$ don't intersect and are within the range $-1 \le y \le 1$

(d) ... a cosine graph and a tangent graph which meet the x-axis the same number of times between x=−4 and x=4?  What if these points have to be the same for both graphs?

Plotting a simple $\tan{x}$ and $\cos{x}$ graph we see that one has 3 and the other has 2 x-axis intercepts in the range $-4 \le x \le 4$.

The easiest and most immediate solution you think of is translating the cosine graph to a sine graph equivalent, having three intersects in that domain.

e.g.

$y=\cos{(x-\frac{\pi}{2})}$ and $y=\tan{x}$ both have exactly three intersections in $-4 \le x \le 4$ and they are at the same points, $x=-\pi,0,\pi$.

Since the tan graph is derived from a sine and a cosine graph, $\tan{x}=\frac{\sin{x}}{\cos{x}}$, the x-intersections of the tan graph are the same as the sin graph. So the only way of getting the same intersections is transforming the cos function into a sin one by a horizontal translation.