Sketching graphs of trig functions. Please help :(

The points of interest will be, intersections with the axes and maximum and minimum points
For example $\cos\frac{2}{3}\theta=0 \mathrm{ \ when\ }\frac{2}{3}\theta=\pm90^o \Rightarrow \theta=\pm135^o$

Thanks for the links. After doing the transformation, what method do i use to find the co-ordinates of interest? Is it not possible to also find these points without doing the sketch?

Points of intersection, max and min points. For example,
Cos2theta/3 = 0
Arcos(0) = 2theta/3 = 90
So theta = 90*3/2
Then use the periodicity if the graph to find the other values of theta that equal 0 in the region -120< x <120

Yes it is possible to do without sketching the graph, but I think whilst you are still learning you should sketch the graphs each time. Will help you understand transformations and also will help you see how many points there are to find.

For example, would you have a rough idea of how many maximum and minimum points to expect for cos(3x), cos(0.5x) or cos(2x) without sketching the graph?
Or could you easily sketch the graph of y= 3cos(3x) + 3 ?
If you need the practice it would be a good idea to draw them. Eventually it will become second nature

But how do i find out which values to substitute?

Thank you again. Yes you are right, i need more practice drawing them. Even now i am still confused about how to do it. When you say the periodicity of the graph, is this for the new graph, after "2/3" has affected it? I am not sure i have it completely right. Have you worked out the values for q A? Would be great if you could post them so i can see how it's done working backwards, and then i'd be more confident doing B too. Thanks for the help thus far though

Sorry, I was busy for the past few hours! To be honest question B is the easier one, it would be a better one to start with

Ok so by periodicity I mean that the graph repeats itself every few degrees.
Think of normal cos. it crosses the y axis at 90, 270, -90 and -270.
+/- 270 is outside your range, and as the graph has been stretched by a factor of 3/2, these areas will have not moved into the range (if the graph had been squashed instead, then we may need to consider values from outside the range)

So I've already mentioned earlier how to show that 2theta/3 = 90, which by rearranging gives theta = 135.
This can also be used for -90, which gives theta = -135.
So you have points of (135, 0) and (-135, 0).

Thanks very much. So it's just 4 values for A? I do understand how you worked that out now. So thank you. Funnily enough, Question B seems harder for me. Even though theta isn't affected, i'm still confused at what the correct values are and how to find them

For question B, the points where the graph crosses the axis will be different.

Best way to work this out is algebraically:

2sinx -1 = 0
2sinx = 1
Sinx = 1/2
Arcsin(1/2) = x

X = 30

Now the periodicity of the graph can be used to work out where the other points of axes intersection are.
As the graph has shifted downwards, only point between 0 and 180 will cross the x axis.

The graph is symmetrical, so if it crosses the x axis at (30,0), then it also crosses at (150,0) under this transformation. (0 to 30 is the same distance as 180 - 30)

How do you work out 135?