On the same set of axes, sketch and label the
graphs of the equations
y = 2cos(2x)
and
y = -2sin(x)
in the interval 0 < x < .
find the values which 2 cos 2x = -2 sin x
The rule for the graphs of:
y = Asin(Bx) and y = Acos(Bx)
both have these important points on the x-axis:
0, ± where n is any integer.
The peaks all have y-coordinate |A|.
The valleys all have y-coordinates -|A|
If A is positive,
y = Asin(Bx) circulates from x-intercept at the origin, peak, x-intercept, valley, x-intercept, peak, etc.
y = Acos(Bx) circulates from a peak at the origin, x-intercept, valley, x-intercept, peak, etc.
Peaks and valleys in the rules above reverse if A is negative.
So for y = 2cos(2x)
A = 2, B = 2. This graph has important points at
x = 0, , , ,
which simplify to
x = 0, , , ,
It has
a peak at (0,2),
an x-intercept at (,0),
a valley at (,-2),
an x-intercept at (,0),
and a peak at (,2)
This is the graph:
Now for y = -2sin(x)
A = -2, B = 1. This graph has important points at
x = , , , ,
which simplify to
x = , , , ,
but only the first three are within the interval 0 < x <
Since A is negative, the peaks and valleys switch from the
rules above and there are only these on the interval 0 < x <
an x-intercept at (0,0),
a valley at (,-2),
an x-intercept at (,0)
This is the graph (it only has one arch on this
interval 0 < x < .
Now let's put them both on the same set of axes:
To find the solution of
2 cos 2x = -2 sin x
we just observe that the solution will be the x-coordinate of
the point where the two curves intersect.
Well it looks as though they intersect only at one point,
which is (,-2), their common valley point.
So the solution in the interval 0 < x < is x =
Edwin