Equation in rectangular coordinates of the quadric surface consisting of the two cones φ = π /4 and Φ = 3π/4?

These are the two halves of the full cone with a pi/4 angle from the cone's axis along the z-axis, and the surface of the cone. The first equation is the z >= 0 half of the cone, and the second is the z <= 0 half.

The full equation for a cone with an angle φ between the axis, the axis on the z-axis, and vertex at the origin is:

x^2 + y^2 = (z*tan φ)^2

Since tan pi/4 = 1, your full cone is:

x^2 + y^2 = z^2

o.k. then. tan(?)=4/3, suitable? attempt squaring the two factors of the equation to get a clean equality. tan^2(?)=sixteen/9 is the end result. Now, replace tan^2(?) in accordance to the hinted equation: tan^2(?)=sin^2(?)/cos^2(?) (that's purely the squared variety of tan=sin/cos) , to get sin^2(?)/cos^2(?)=sixteen/9 , which now needs to be rewritten. Cos^2(?)+sin^2(?)=a million, by potential of a few arcane info you probable discovered at college. meaning that Cos^2(?)=a million-sin^2(?) and sin^2(?)=a million-cos^2(?) , yet purely the 1st concerns right here. replace cos^2(?) with a million-sin^2(?) interior the above ( sin^2(?)/cos^2(?)=sixteen/9 ) for Sin^2(?)/(a million-sin^2(?))=sixteen/9. Now to sparkling up for sin^2(?), and then sin(?). Multiply the two factors of this by potential of a million-sin^2(?) , with a view to have Sin^2(?)=(sixteen/9)*(a million-sin^2(?)). boost the suitable area . . Sin^2(?)=sixteen/9 - (sixteen/9)*sin^2(?) . . and upload the sixteen/9 sin^2(?) to the left: (9/9+sixteen/9)*sin^2(?)=sixteen/9 , which simplifies to twenty-5/9*sin^2(?)=sixteen/9 . . and that i think of you could divide the two factors by potential of 25/9, then locate the sq. root of the two one among the two factors on your person.