Can someone please help me with this trig/3-dimensional geometry problem...?

Question

I'm supposed to show that for all values of j and k, the point (a sink cosj, a sink sinj, a cos k) lies on the sphere x^2 +y^2 + z^2 = a^2.

If I plug the coordinates into the equation I get: (a sink cosj)^2 + (a sink sinj)^2 + (a cosk)^2 = a^2

After that I don't know what to do; please help!

Answer 1

You're actually on the right track - let me show you where to go from where you left off:

(a sin k cos j)^2 + (a sin k sin j)^2 + (a cos k)^2 = a^2

(a^2 sin^2 k cos^2 j) + (a^2 sin^2 k sin^2 j) + (a^2 cos^2 k) = a^2

[a^2] (sin^2 k cos^2 j) + (sin^2 k sin^2 j) + (cos^2 k) = a^2

(sin^2 k cos^2 j) + (sin^2 k sin^2 j) + (cos^2 k) = 1

(sin^2 k cos^2 j) + (sin^2 k sin^2 j) = 1 - (cos^2 k)

[sin^2 k] (cos^2 j + sin^2 j) = 1 - (cos^2 k)

sin^2 k = 1 - (cos^2 k)

sin^2 k + cos^2 k = 1