2道三角函数的化简
问题描述:
2道三角函数的化简
(1)sinacosa(tana+cota)
(2)cot^2a(tan^2a-sin^2a)
答
⑴sinacosa(tana+cota)=sinacosa(sina/cosa+cosa/sina)=sinacosa*sina/cosa+sinacosa*cosa/sina=sin^2a+cos^2a=1⑵cot^2a(tan^2a-sin^2a)=cot^2a*tan^2a-cot^2a*sin^2a=1-cos^2a*sin^2a/sin^2a=1-cos^2a=sin^2a...