求证sina^2+sin&^2-sina^2*sin&^2+cosa^2*cos&^2=1

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/21 10:22:05
求证sina^2+sin&^2-sina^2*sin&^2+cosa^2*cos&^2=1
xM0ʴ4H`]rDN[#\+8Sf7]u?5*hTР*KtyPq륧 waIX#"Y);QvD8>M?h+0H&I%0N$yl /tD2(]r2dg}B7~Fw)eB~iqF$lv(3[Jah

求证sina^2+sin&^2-sina^2*sin&^2+cosa^2*cos&^2=1
求证sina^2+sin&^2-sina^2*sin&^2+cosa^2*cos&^2=1

求证sina^2+sin&^2-sina^2*sin&^2+cosa^2*cos&^2=1
原式=sina^2+(sin&^2-sina^2*sin&^2)+cosa^2*cos&^2
=sina^2+sin&^2(1-sina^2)+cosa^2*cos&^2
=sina^2+(sin&^2*cosa^2+cosa^2*cos&^2)
=sina^2+(sin&^2+cos&^2)*cosa^2
=sina^2+cosa^2
=1

(sina)^2+(sin&)^2-(sina)^2*(sin&)^2+(cosa)^2*(cos&)^2
=(sina)^2) + (sin&)^2(1-(sina)^2) + (cosa)^2(cos&)^2)
=(sina)^2 + (sin&)^2(cosa)^2 + (cosa)^2(cos&)^2
=(sina)^2 + (cosa)^2
=1.

sina=sin[(a+b)/2+(a-b)/2]=sin(a+b)/2cos(a-b)/2+cos(a+b)/2sin(a-b)/2 sinb=sin[(a+b)/2-(a-b)/2]=sin(a+b)/