sin^2a+sin^2β-sin^2a*sin^2β+cos^2a*cos^2β=1 证明恒等式?各位高手 教教我

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/27 23:51:40

sin^2a+sin^2β-sin^2a*sin^2β+cos^2a*cos^2β=1 证明恒等式?各位高手 教教我
sin^2a+sin^2β-sin^2a*sin^2β+cos^2a*cos^2β=1 证明恒等式?各位高手 教教我

sin^2a+sin^2β-sin^2a*sin^2β+cos^2a*cos^2β=1 证明恒等式?各位高手 教教我
证明:sin^2a+sin^2β-sin^2a*sin^2β+cos^2a*cos^2β
=sin^2a+sin^2β-sin^2a*sin^2β+(1-sin^2a)(1-sin^2β)
=sin^2a+sin^2β-sin^2a*sin^2β+1-sin^2a-sin^2β+sin^2a*sin^2β
=1

sin^2a+sin^2β-sin^2a*sin^2β+cos^2a*cos^2β=sin^2a(1-sin^2β)+sin^2β+cos^2a*cos^2β=sin^2acos^2β+cos^2a*cos^2β+sin^2β=cos^2β(sin^2a+cos^2a)+sin^2β=cos^2β+sin^2β=1