cos3q = sinq
By Arun Iyer on Friday, November 23, 2001
- 06:30 pm:
In a particular mechanics problem,
i ran into a equation as such...
cos3q = sinq
is there any easy method of solving this equation....??
i actually solved it by taking cosq = x
which gives...
i.e. x6+x2-1=0
then i took f(x)=x6 +x2 -1 and solved for x
by Newton Raphson method....
this was quite tedious and time consuming especially during
exam....
hence i need any easier method if possible...
love arun
By Yatir Halevi on Friday, November 23,
2001 - 07:16 pm:
actually there is a way,
take x2 =m
then we have
m3 +m-1=0
And this is the depressed for of the cubic expression (without
the x2 ), and there is a simple way of solving this,
described in:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
Hope it helps,
Yatir
By Andrew Hodges on Friday, November 23,
2001 - 07:18 pm:
f(x) is a depressed cubic in x2 if that helps you
Arun...
By Kerwin Hui on Saturday, November 24,
2001 - 11:42 am:
Arun,
Here is a quicker way to get to the cubic. First, notice that we cannot have
cosq = 0, so divide both sides by cos3q and we obtain
1=tanq+tan3q
Now we just plug this into del Ferro's, or use some other methods.
Kerwin
By Arun Iyer on Saturday, November 24,
2001 - 06:04 pm:
Well, thanks for all the methods suggested...
though, I would like to have some quicker method to solve the
real equation...i.e
cos3q = sinq
love arun