Question

高一三角函数恒等变换，求过程

已知cos(π/4-α)=12/13, 且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)

Answer 1

解答：
因为：cos(π/4-α)=12/13，所以：cosπ/4cosα+sinπ/4sinα=12/13，
即：（根号2/2)(sinα+cosα)=12/13，所以：sinα+cosα=(12倍根号2)/13
两遍呢同时平方得到：1+sin2α=288/169，解得：sin2α=119/169
因为：π/4-α是第一象限角 ，所以：2kπ<π/4-α<2kπ+π/2
得到：2kπ-π/2<α-π/4<2kπ，解得：2kπ-π/4<α<2kπ+π/4
所以：4kπ-π/2<2α<4kπ+π/2，即：2α为第一或第四象限角
所以：sin(π/2-2α)/sin(π/4+α)=cos2α/cos[π/2-(π/4+α)]=根号下[1-(sin2α)^2]/cos(π/4-α)
=120/169*13/12=10/13

Answer 2

由cos(π/4-a)=12/13,且π/4-α是第一象限角
得sin(π/4-a)=5/13,
从而sin(π/2-2α)/sin(π/4+α)
=sin2(π/4-α)/sin(π/4+α)
=2sin(π/4-α)cos(π/4-α)/sin[π/2-(π/4-α)] (注：π/4+a=π/2-(π/4+a)),
=2sin(π/4-α)cos(π/4-α)/cos(π/4-α)
=2sin(π/4-α)
=2*(5/13)
=10/13

Answer 3

已知cos(π/4-α)=12/13, 且π/4-α是第一象限角 ,则sin(π/2-2α)/sin(π/4+α)＝？
解：sin(π/2-2α)/sin(π/4+α)=cos2α/[(√2/2)(cosα+sinα)]=(cos²α-sin²α)/[(√2/2)(cosα+sinα)]
=(√2)(cosα-sinα)=2[cos(π/4)cosα-sin(π/4)sinα]=2cos(π/4+α)=2cos[π/2-(π/4-α)]=2sin(π/4-α)
=2√[1-cos²(π/4-α)]=2√[1-(12/13)²]=10/13