计算4tan²π/4-cos²π/3-1/2sin²π/6+sinπ/2*cos0的值
问题描述：
计算4tan²π/4-cos²π/3-1/2sin²π/6+sinπ/2*cos0的值
答
原式=4×1²-(1/2)²-1/2×(1/2)²+1×1
=4-1/4-1/8+1
=37/8