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 $sin^8(x)+cos^8(x)=(sin^2(x)+cos^2(x))sin^{8}(x)+(sin^2(x)+cos^2(x))cos^{8})(x)=sin^{10}(x)+cos^{10})(x)
==Solution 2 (Another BASH)==
First, for simplicity, let$ (Error compiling LaTeX. Unknown error_msg)a=\sin{x}
b=\cos{x}
a^2+b^2=1
a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1
\frac{11}/{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)
a^2b^2(a^8+b^8)
y=a^2b^2
a^2+b^2=1
a^6+b^6+3a^2b^2(a^2+b^2)=1
a^6+b^6=1-3y
\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1
a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}y(1-3y)+2y^2=y-y^2=\frac{5}{36}
y=1y=\frac{1}{6}
y=1
y=\frac{1}{6}
a^4+b^4=(a^2+b^2)-2a^2b^2=\frac{2}{3}
a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}
a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}*\frac{1}{6}=\frac{13}{54}
\boxed{067}$.
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 | ||
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