Hi there im in my final year at high school in Scotland and i am having difficulty with the wave function.
for example in the exercise we are told to write the expression :
a cos x+ b sin x in the form k cos(x-a)
This i had little problem with and was able to work out what quadrant i shoul use etc what i mean is the
all positive. sin positive, tan positive and cosine positive quadrants
but the next exercise was expressing
a cos x + b sin x in other forms such as:
k cos(x+a) k sin(x-a) and k(sin x+a)
with these i am able to work out the k. however when it comes down to the quadrat i do not no whether to take away 180 add 180 or take away 360 if you no what i mean.
April 17th 2005, 08:38 PM
a = Acos(alpha)
b = Asin(alpha)
acos(x) + bsin(x) = Acos(x-alpha)
If you've already written it in the form k cos (x-a), then you can get the others simply by doing simple arithmetic, or applying basic facts about the trig functions...
Or, isn't it possible to do the same type of work to arrive at that form from scratch? ...
April 21st 2005, 05:17 AM
There is NO "quadrant" involved! x can by anything so this can be in any quadrant.
The basic rule you want is the "sum theorem": cos(u- v)= cos(u)cos(v)+ sin(u)sin(v).
In order to convert a cos x+ b sin x to k cos(x- phi) (NOT "x-a"- you are already using a as the coefficient of cos x!) , you need to take u= x and then find v= phi so that cos(v)= a, sin(v)= b. Of course, that probably isn't possible for arbitrary a and b: you must have a^2+ b^2= 1. To fix that, multiply and divide by a^2+ b^2: (a^2+ b^2)(a/(a^2+b^2) cos x+ b/(a^2+ b^2)sin x). Okay, now it is obvious that you want k= (a^2+ b^2) and you need to find phi such that sin(phi)= a/(a^2+ b^2) (for specific a and b, you may need to use a calculator for that).
April 21st 2005, 11:08 AM
a cos(x) + b sin(x) = k cos(x -A)
At x = 0,
a cos(0) + b sin(0) = k cos(0 -A)
a(1) +b(0) = k cos(-A)
a = k cosA -------(1)
(You know cos(-A) = cosA also.)
At x = 90 degrees,
a cos(90deg) + b sin(90deg) = k cos(90deg -A)
a(0) +b(1) = k sinA
(You know cos(90deg -A) = sinA also.)
b = k sinA -----(2)
With those Eq.(1) and Eq.(2) you can solve for k and angle A.
(2) divided by (1),
b/a = (k sinA)/(k cosA)
b/a = sinA/cosA
b/a = tanA
A = arctan(b/a) ----answer.
Then, with A already known, say, in (1),
a = k cosA
k = a/cosA ----answer.