作业帮计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)],
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 12:19:33
![计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)],](/uploads/image/z/10264668-60-8.jpg?t=%E8%AE%A1%E7%AE%974%2F3%5Bcos%5E2%2813%CF%80%2F6%29%5D%2B3tan%5E2%28%CF%80%2F6%29-1%2F3%5Bsin%5E2%282%CF%80%2F3%29%5D%2C)
计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)],
计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)],
计算4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)],
(4/3)*cos²(13π/6)+3tan²(π/6)-(1/3)*sin²(2π/3)
=(4/3)*cos²(π/6)+3tan²(π/6)-(1/3)*sin²(π/3)
=(4/3)*(√3/2)²+3*(√3/3)²-(1/3)*(√3/2)²
=(4/3)*(3/4)+3*(1/3)-(1/3)*(3/4)
=1+1-1/4
=7/4
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4/3[cos^2(13π/6)]+3tan^2(π/6)-1/3[sin^2(2π/3)],
=4/3cos^2π/6+3tan^2π/6-1/3sin^2π/3
=4/3*3/4+3*1/9-1/3*3/4
=1+1/3-1/4
=13/12
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