高校问答信用卡逾期还能分期吗:化简一道三角函数
来源：百度文库 编辑：高校问答 时间：2024/04/29 18:28:22
[tg(π-φ)/sin(π-φ)cos(-φ-π)]-[sin(φ-2π)sin(3π+φ)/cos(5π-φ)cos(φ-2π)]
解:[tg(π-φ)/sin(π-φ)cos(-φ-π)]-[sin(φ-2π)sin(3π+φ)/cos(5π-φ)cos(φ-2π)]
=[-tgφ/sinφcos(φ+π)]-[-sin(-φ+2π)sin(3π+φ)/cos(5π-φ)cos(-φ+2π)]
=[-tgφ/sinφ(-cosφ)]-[-sinφ(-sinφ)/(-cosφ)cosφ]
=[(sinφ/cosφ)/sinφcosφ]+[sinφsinφ/cosφcosφ]
=1/cos^2φ+tan^2φ
=sec^2φ+tan^2φ
=2tan^2φ+1