How to get θ sin and cos are together in 1 equation

[ohaiyo88]

b²cos²θ+2cosθ(-mb+bc²-bz-bcsinθ)+2sinθ(cm-cz)-c²sin²θ=p

How to get θ in terms of b,c,m,z,p
The equation i formed to calculate the dynamic
PLs, anyone can help to solve this, i would much appreciate! I am doing my FYP!

 

[HallsofIvy]

Replace sin^2(\theta) with 1- cos^2(\theta) and sin(\theta) with \sqrt{1- cos^2(\theta)}. That will give you an equation in cos(\theta) only. However it will involve a square root so isolate that on one side of the equation and square both sides to get a fourth degree equation in cos(\theta). That is not going to give a simple answer!

 

[ohaiyo88]

Thanks for ur precious reply! Somehow i still have to go through this question.. does it has any alternative method?

 

[phyzguy]

HallsofIvy's approach is the right one. Do you know the values of b,c,m,z,p (or ranges)? The equation will be easily solved numerically if you do.

 

[ohaiyo88]

b,c,m,z,p are the fixed constant, they will be known later. Somehow i can't get θ if i use HallsofIvy's approach because the existence of fourth, third, second, first degree of cos θ

 

[ohaiyo88]

and also how if i want to use numerical method? how to use it ? Matlab?

 

[phyzguy]

Personally, I use Mathematica, but Matlab will work. A simple Newton's method solver will find the root quickly once the coefficients are known.

 

[ohaiyo88]

may i know how do i use numerical method to solve this trigo+algebraic equation? i need to equate the equation θ in terms of b,c,m,z,p

Using Newton Raphson? (i donno how to use software, only know how to use handwritten)

 

[phyzguy]

You could try Wolfram Alpha, which uses the Newton_Raphson method in the "FindRoot" function. For example, when I took the values, {b=1, c=2, z=3, m=4, p=5}, then using x for cos(theta), the equation becomes:

(-5 - 6 x + x^2 - 4 (1 - x^2))^2 == (-4 + 4 x)^2 (1 - x^2)

Wolfram Alpha, using the link below, finds x = -.45969. you can then find theta using the arccos function, or a calculator.

http://www.wolframalpha.com/input/?...-4+(1-x^2))^2+==+(-4+4+x)^2(1-x^2),+{x,+0.5}]