---------------============ 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
Last edited by Peritus; January 14th, 2008 at 02:58 PM.
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