Derive a trigonometric equation for the volume of the cone

[Atomos]

A circular cone is inscribed in a sphere with a radius of 30cm. The semi vertical angle is theta. Derive a trigonometric equation for the volume of the cone.

This has be stumped. I tried looking up proofs for the expression of the volume of a cone for inspiration but all involve calculus.

 

[Atomos]

[Y]es this can be solved with pre-calculus you fool, or [N]o you cannot solve this without ising calculus

 

[mezarashi]

You don't need calculus for this. Calculus is the mathematics of change. If the question were to ask something like, the radius of the sphere was changing by this much, how fast is the height of the cone changing, THEN calculus would be needed.

However, in this case, you need to find the relationship between the cone's dimensions and the sphere's radius... which WOULD BE a prerequisite to a calculus problem if it were to be.

 

[Tide]

[M]aybe! You can do it with sums (volumes of thin disks) and you don't have to call it calculus! :)

 

[stmoe]

the radius of the cone's base depends on the position it is inside of the sphere... I would start by looking at it 'from the side' .. as just an isosceles triangle inside a circle, then you can use the equation of a cirlcle to figure out how the triangle's base will compare... then start looking at the actual volumes of ,

.. you can do it without calculus ... .. but then again, if I don't think I would unless I absolutely had to.

 

[Atomos]

[M]aybe! You can do it with sums (volumes of thin disks) and you don't have to call it calculus! :)

This solution came to me as soon as I saw the question, however, I am not "supposed" to know any type of mathematics regarding the sum of an infinite number of infinitesimal changes.

Stmoe: I could make the problem as simple as just finding the height and width in terms of the radius of the circle and vertex angle and then sub these values into my cone volume formula, however, I am not sure if the questions wants me to derive that formula as well.