2019 AIME I Problems/Problem 8

Revision as of 19:39, 14 March 2019 by Mathboy11 (talk | contribs) (Solution)

The 2019 AIME I takes place on March 13, 2019.

Problem 8

Solution

Remember $sin^2(x)+cos^2(x)=1$. This means $sin^{10}(x)+cos^{10}(x)=(sin^2(x)+cos^2(x))sin^{10}(x)+(sin^2(x)+cos^2(x))cos^{10})(x)=sin^{12}(x)+cos^{12}(x)+sin^2(x)cos^{10}(x)+sin^{10}(x)cos^{2}(x)$

See Also

2019 AIME I (ProblemsAnswer KeyResources)
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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