问题
解答题
(1)已知
(2)计算:sin230°+cos245°+
|
答案
(1)原式可变形为:7(2a-b)=4(a+2b)
整理得:10a=15b
故:
=a b
=15 10 3 2
(2)原式=(
)2+(1 2
)2+2 2
×2
×13 2
=
+1 4
+2 4 6 2
=
+3 4 6 2
(1)已知
(2)计算:sin230°+cos245°+
|
(1)原式可变形为:7(2a-b)=4(a+2b)
整理得:10a=15b
故:
=a b
=15 10 3 2
(2)原式=(
)2+(1 2
)2+2 2
×2
×13 2
=
+1 4
+2 4 6 2
=
+3 4 6 2