tanx=2求2/3sinx²+1/4cosx²
问题描述:
tanx=2求2/3sinx²+1/4cosx²
答
tanx=2求
2/3sinx²+1/4cosx²=(2/3sinx²+1/4cosx²)/(sinx²+cosx²)=[(2/3sinx²+1/4cosx²)/cosx²]/[(sinx²+cosx²)/cosx²]=[2/3tanx²+1/4]/[tanx²+1]=[2/3*2²+1/4]/[2²+1]=[8/3+1/4]/5=