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.
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date: 2021-08-09 16:56
summary: when is sin^8(x) + cos^6(x) = 1?
tags:
- '@?maths'
- '@?mat'
- '@?notes'
- '@?public'
title: MAT - Paper 2011 - Q1I