求函數y=3Sin(x+20')+5Sin(x+80')的最大值

熱心網友

y=3sin(x+20)+5sin(x+80),本題中略去角度制單位:°.=3(sinxcos20+cosxsn20)+5(sinxcos80+cosxsin80)=(3cos20+5cos80)sinx+(3sin20+5sin80)cosx∵形如y=asinx+bcosx=√(a^2+b^2)sin(x+φ)的函數的最大值是√(a^2+b^2)∴y(大)=√[(3sin20+5sin80)^2+(3cos20+5cos80)^2]化簡,(3cos20+5cos80)^2+(3sin20+5sin80)^2=9+25+30(cos80cos20+sin80sin20)=34+30cos(80-20)=34+30cos60=34+15=49∴原函數最大值y=√49=7