MABCD is a regular pyramid (ABCD is square and the lateral edges are equal). The angle between the base ABCD and the plane through BD, which is perpendicular to MC, is $\phi$. Then what is the sine of the angle between MB and the plane MAD.
I have found it to be $sin2\phi\over \sqrt{1+cos^2\phi} $, but using very tedious approach involving a lot of calculations. (expressing all the edges with $\phi$ and the lateral edge, translating MB and applying the triangle similarity principles, the Pythagorean and Cosine Theorems a few times).
Is my answer correct and is there an easier approach?
It can be done a lot easier using altitudes and volumes.