I did the wrong homework sheet, so now I have to have this sheet finished tonight. I was off school for a few days so I missed some of the explanations given on how to do everything, and I'd appreciate some help. It's three questions I'm stuck on... I'm sure they'll look easy to other people; I've seen topics on here what are wayy advanced to what I'm doing. Sorry, I'm rambling. Here're the questions.







Thanks so much if you can help! I was really worried before I found this forum. >_<

---------------============ Q3 ===========----------

(a)

as you can see this function looks like the sin function although it has adifferent period length, it repeats itself every 120 instead of 360, also it's amplitute is 4 instead of 1, thus a suitable function would be 4*sin(3*x).

(b) this is just the tangent function -> tan(x)

(c) this function looks like the cos function although it has a shorter period,a bigger amplitude and an opposite phase
-> f(x) = -8*cos(4*x)

---------------============ Q9 ===========----------
if f(x) = f(x+p) for all x then the function is p periodic, following this principle we can solve (a):

3*cos(2*x) = 3*cos[2*(x+p)]
<=> 2*x = 2*(x+p) + 2*pi (cos(x) has a period of 2*pi)

p = pi

I'm sure you'll be able to do the rest of the question...

---------------============ Q15 ===========----------
(a) 8*tan(x) - 3 = 2

<=> tan(x) = 5/8

x = atan(5/8) + pi*k [remember that tan is pi periodic]

(b)...

(c) 4*cos(x)^2 - 1 = 0

<=> 2*cos(x)^2 = 1/2

using the double angle identity -> 2*cos(x)^2 - 1 = cos(2*x)

we get: cos(2*x) + 1 = 1/2

<=> cos(2*x) = -1/2

2x = (2/3)*pi + 2*pi*k ---> x = (1/3)*pi + pi*k
and also:

2x = (4/3)*pi + 2*pi*k ---> x = (2/3)*pi + pi*k