Problem:
Given that log10sinx+log10cosx=−1 and that log10(sinx+cosx)=21(log10n−1), find n.
Solution:
Use logarithm properties to obtain log(sinxcosx)=−1, and then sinxcosx= 1/10. Note that
(sinx+cosx)2=sin2x+cos2x+2sinxcosx=1+102=1012
Thus
2log(sinx+cosx)=log1012=log12−1
so
log(sinx+cosx)=21(log12−1)
and n=12.
The problems on this page are the property of the MAA's American Mathematics Competitions