共回答了1627个问题 举报
(tanα)^2=1/2
tanα = 1/√2 or -1/√2
cos2α = 2(cosα)^2 -1
= 2(√2/3)^2 -1
= 4/9 - 1
= -5/9
(sin2α)^2 = 1-(cos2α)^2
= 1- 25/81
= 56/81
sin2α = 2√14/9 or -2√14/9
tanα = 1/√2 or -1/√2
cos2α = 2(cosα)^2 -1
= 2(√2/3)^2 -1
= 4/9 - 1
= -5/9
(sin2α)^2 = 1-(cos2α)^2
= 1- 25/81
= 56/81
sin2α = 2√14/9 or -2√14/9
1年前
0
可能相似的问题
-
已知tanα=-3/4,α在第二象限,求cos2α+sin2α
1年前1个回答
-
（1）已知tanθ=2，求[1+sin2θ/cos2θ]的值；
1年前1个回答
你能帮帮他们吗