作业帮高一关于三角恒等变形的题目
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/05/01 04:54:16
高一关于三角恒等变形的题目
已知α和β都是锐角,且sinα=asinβ,tanα=btanβ求证cosα=√a^2-1/b^2-1
楼下这位仁兄怎么搞的?
已知α和β都是锐角,且sinα=asinβ,tanα=btanβ求证cosα=√a^2-1/b^2-1
楼下这位仁兄怎么搞的?
sinα=asinβ
两边平方
(sinα)^2=a^2(sinβ)^2
(sinβ)^2=(sinα)^2/a^2
所以(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2
所以(tanβ)^2=(sinβ)^2/(cosβ)^2=(sinα)^2/[a^2-(sinα)^2]
所以(tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/[a^2-(sinα)^2]
(cosα)^2=(sinα)^2/(tanα)^2=[a^2-(sinα)^2]/b^2
b^2(cosα)^2=a^2-(sinα)^2=a^2-[1-(cosα)^2]=a^2-1+(cosα)^2
(cosα)^2=(a^2-1)/(b^2-1)
是锐角
cosα>0
所以cosα=√[(a^2-1)/(b^2-1)]
两边平方
(sinα)^2=a^2(sinβ)^2
(sinβ)^2=(sinα)^2/a^2
所以(cosβ)^2=1-(sinβ)^2=[a^2-(sinα)^2]/a^2
所以(tanβ)^2=(sinβ)^2/(cosβ)^2=(sinα)^2/[a^2-(sinα)^2]
所以(tanα)^2=b^2*(tanβ)^2=b^2(sinα)^2/[a^2-(sinα)^2]
(cosα)^2=(sinα)^2/(tanα)^2=[a^2-(sinα)^2]/b^2
b^2(cosα)^2=a^2-(sinα)^2=a^2-[1-(cosα)^2]=a^2-1+(cosα)^2
(cosα)^2=(a^2-1)/(b^2-1)
是锐角
cosα>0
所以cosα=√[(a^2-1)/(b^2-1)]