Good questions. See my notes below and let me know if they don't answer you question.
Sir,
For the following question, y=4tan 1/2x ...I am having difficulty constructing this graph...
Bearing in mind there are asymptote at each odd multiple of Pi/2 3pi/2 5pi/2 etc, because the value of the y=(sin Theta/cos Theta) is undefined at these points.. I plotted four rows..laid out values of x ( o, pi/6, pi/4 ,pi/3) etc, in the first , In the second row, I put ( 1/2 * x) values. In the third, I put y= tan x values and in the fourth I have y=4* tan (1/2x).
Now, I know the period of the function is 4pi and that the values for (y) will be stretched out, but I'm not sure where my asymptotes should be and how to determine them either...Can you please advise..
The period of tan(theta) is pi (not 2 pi as it would be for sine or cosine).
If theta = 1/2 x then x = 2 * theta and the period becomes 2 pi (not 4 pi).
Asymptotes occur for theta = pi/2, 3 pi/2, etc. so they will occur for x values 2 * pi/2, 2 * 3 pi / 2, etc., i.e., for pi, 3 pi, etc..
The following is the strategy used in qa #2:
I recommend laying out the columns for theta and tan(theta) first, as in qa 2. Your theta values will be 0, pi/6, pi/4 etc.
Then setting 1/2 x = theta you solve for x to get x = 2 * theta. This would give you x values 0, pi/3, pi/2, etc.
You would end up with something like this
x theta = 1/2 x tan(theta) 4 tan(theta)
0 0 0 0
pi/3 pi/6 2/sqrt(3) 8 / sqrt(3)
pi/2 pi/4 1 4
pi pi/3 sqrt(3)/2 2 sqrt(3)
...
Your asymptotes would correspond to theta = pi/2, 3 pi/2, etc., and therefore to x = pi, 3 pi, etc..
The period would therefore be 2 pi, as it should.
Theta intervals of pi/6 become x intervals of 2 * pi/6 = pi / 3.