已知tanx=2,求(2⼀3)sin눀x+(1⼀4)cos눀x的值

2025-06-28 11:44:32
推荐回答(1个)
回答1:

tanx=2
(2/3)sin^2(x)+(1/4)cos^2(x)
=[(2/3)sin^2(x)+(1/4)cos^2(x)]/1
=[(2/3)sin^2(x)+(1/4)cos^2(x)]/[sin^2(x)+cos^2(x)]
分子分母同除以cos^2(x)得:
原式=[(2/3)tan^2(x)+(1/4)]/[tan^2(x)+1]
=[(2/3)*4+(1/4)]/[4+1]
=[(8/3)+(1/4)]/5
=7/12