Difference between revisions of "2019 AIME I Problems/Problem 8"
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==Problem 8== | ==Problem 8== | ||
==Solution== | ==Solution== | ||
| − | Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>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)= | + | Remember <math>sin^2(x)+cos^2(x)=1</math>. This means <math>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}[sin^8(x)+cos^8(x)</math>. |
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=7|num-a=9}} | {{AIME box|year=2019|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:42, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 8
Solution
Remember 
. This means 
.
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. 
