$\cos^{3} θ = \sin θ$

By Arun Iyer on Friday, November 23, 2001 - 06:30 pm:

In a particular mechanics problem,
i ran into a equation as such...

\cos^{3} θ = \sin θ

is there any easy method of solving this equation....??

i actually solved it by taking

\cos θ = x

which gives...

$x^{3} = \sqrt{1 - x^{2}}$

i.e. $x^{6} + x^{2} - 1 = 0$
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 $\cos θ = 0$ , so divide both sides by $\cos^{3} θ$ and we obtain

$1 = \tan θ + \tan^{3} θ$

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

\cos^{3} θ = \sin θ

love arun