反三角函数求角度
2个回答
展开全部
解:假设所求角为锐角(因为三角函数为周期函数,这样能少不少步骤)
令arctan(2+√3)=α(0<α<π/2)
则tanα=2+√3
即tanα=(2+√3)/1
∴cosα=1/√[(2+√3)²+1²]
=1/√(8+4√3)
=1/√(√6+√2)²
=1/(√6+√2)
=(√6-√2)/(√6+√2)(√6-√2)
=(√6-√2)/4
=(√3/2)×(√2/2)-(1/2)×(√2/2)
=(cosπ/6)×(cosπ/4)-(sinπ/6)×(sinπ/4)
=cos(π/6+π/4)
=cos5π/12
∴ α=5π/12
令arctan(2+√3)=α(0<α<π/2)
则tanα=2+√3
即tanα=(2+√3)/1
∴cosα=1/√[(2+√3)²+1²]
=1/√(8+4√3)
=1/√(√6+√2)²
=1/(√6+√2)
=(√6-√2)/(√6+√2)(√6-√2)
=(√6-√2)/4
=(√3/2)×(√2/2)-(1/2)×(√2/2)
=(cosπ/6)×(cosπ/4)-(sinπ/6)×(sinπ/4)
=cos(π/6+π/4)
=cos5π/12
∴ α=5π/12
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询
广告 您可能关注的内容 |