trig graph questions help please and simplify questions help please?

Question

work please
what is starting point for
1) -cos(2x)
2) -sin(3x)
3) -2 + 4 cos(x)

simplify
4. (-1/64)^(1/2)

Answer 1

4.

Square root of (-1/64) = sqrt(-1)/8 = +-i/8

for 1,2,3 I don't know what you mean by "starting point" but you can start by examining when those functions have zeros and maximum and minimum values

Answer 2

You have prob 4 answered.

With the other problems I assume you want a graph of each trig function you've supplied (?).

1) - cos (2x) You should be acquainted with what the cos x looks like (and also what the sinx looks like) : cosx has an amplitude of 1, and makes a complete cycle in a 2pi interval along the x-axis.

So because of the argument (2x) the interval for a cycle is now 2pi/2 = pi, and the amplitude is unchanged. However, that is a NEG one coefficient, so now the graph of cos (arg) will be reversed; what normally was above the x axis, is now below; what was below will now be above.

So starting at x = 0 and going to x = pi draw a typical cosine curve in 4 equal length quadrants/quarters in the pi interval, BUT instead of starting at y= +1 and going down to zero at x = pi/4, start at y = -1 and go UP to y = zero at x = pi/4. Continue with your inverted cosine curve till you finish the 4th interval at x=pi, y = -1.

2) In a similar fashion for -sin(3x), because of the NEG the sine curve will go down (instead of up) from (0,0) to reach y = -1 when x = (1/4)(2pi/3); then RISE UP to y = 0 when
x = (2/4)(2pi/3). When x = (3/4)(2pi/3) y will have risen to +1; and the cycle will complete at x= 2pi/3 with y again = zero.

3) -2 + 4cosx This is a little more challenging. You have a "regular" cosine function graph, EXCEPT

The amplitude of the graph is 4 (instead of 1). A full cycle will occur in the interval 0<= x <= 2pi; BUT because of the -2 term you no longer use the x axis (where y = 0) as the line of symmetry. Instead you use the line y = -2, and "bounce" the cosx between -6<= y <= 2 (which is 4 up and 4 down from the axis of symmetry, while traversing the zero to 2pi interval for x.

Answer 3

Where do You want to start? generally when I'm looking and graphs of trig functions I think of values 0 some teachers want a graph from -2pi to 2pi.
1) we know that the period of the cos(x) function is 2pi, when you put something in there with the x it changes the behavior. 2x means the function oscillates twice as fast or the period is cut in 1/2 (0 to pi). cosine oscillates between -1 and 1. now we just need to find when that happens.
at x=0, cos(x)=1 so -cos(2x)=-1
and usually x=pi would be 1 but we're looking at the period ending there so this will also be -1.
you know where the period starts and where it ends and how cosine behaves in between, midway thru the period the function=1 and it crosses the x-axis twic in the period.
2) use these same concepts
3) when cos(x)=0 then y=(-2) the period is 2pi use the same concepts to find your intercepts and then smooth connect them.
4) is not a real number because the square root of a negative number is not in the realm of reals so what you really have is
(1/64)^(1/2)times "i"
the following site may be helpful for trig graphs