# MAT - Paper 2011 - Q1I

## Flashcards

When is $y^2 < y$?

For $0 < y < 1$.

Why does $\sin^8(x) + \cos^6(x)$ have to be less than $\sin^2(x) + \cos^2(x)$ for $0 < y < 1$?

Because $\sin(x)$ and $\cos(x)$ are between $1$ and $0$, so squaring them can only make them smaller.

How can you determine the number of solutions to $\sin^8(x) + \cos^6(x) = 1$?

This can only happen when $\sin^2(x) = 1$ and $\cos^2(x) = 0$ or vice versa because of how squaring works.