Question

已知sinα=2cosα，则sin^2α+2sinαcosα=

Answer 1

解：因为 sin α =2 cos α,
所以 tan α =2.

所以 (sin α)^2 +2 sinα cosα =[ (sin α)^2 +2 sin α cos α ] / [ (sin α)^2 +(cos α)^2 ]
=[ (tan α)^2 +2 tan α ] / [ (tan α)^2 +1 ]
=8/5.

= = = = = = = = =
以上计算可能有误！
弦化切，即原式先除以 (sin α)^2 +(cos α)^2，然后分子分母同时除以 (cos α)^2.

= = = = = = = = =
解法2：
令 x =cos α,
则 sin α =2x.
所以 x^2 +(2x)^2 =1,
解得 x^2 =1/5.

所以 原式 =(2x)^2 +2 *2x *x
=8x^2
=8/5.

= = = = = = = = =
换元法。同角三角函数问题，实质上是代数问题。

Answer 2

sin^2α+2sinαcosα=4sinαcosα=4*2cosα^2=8cosα^2