问题
解答题
计算: ①
②tan30°sin60°+cos230°-sin245°tan45°. |
答案
①原式=2
-2×2
+1-32 2
=2
-2
+1-32
=
-2;2
②原式=
×3 3
+(3 2
)2-(3 2
)2•12 2
=
+1 2
-3 4 1 2
=
.3 4
计算: ①
②tan30°sin60°+cos230°-sin245°tan45°. |
①原式=2
-2×2
+1-32 2
=2
-2
+1-32
=
-2;2
②原式=
×3 3
+(3 2
)2-(3 2
)2•12 2
=
+1 2
-3 4 1 2
=
.3 4