2019 AIME I Problems/Problem 8
The 2019 AIME I takes place on March 13, 2019.
Problem 8
Solution(BASH)
Remember 
. This means ![$sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{2}(x)[sin^8(x)+cos^8(x)]$](http://latex.artofproblemsolving.com/5/c/7/5c7fb0d6c7404fe93fcab85035642d9a1da5f540.png)
. Let us look at $sin^8(x)+cos^8(x)=(sin^2(x)+cos^2(x))sin^{8}(x)+(sin^2(x)+cos^2(x))cos^{8})(x)=sin^{10}(x)+cos^{10})(x)
Solution 2 (Another BASH)
See Also
| 2019 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
