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高中数学必修 第一册(648题)



求证:

\({\sin^{2}}\alpha +{\sin^{2}}\beta -{\sin^{2}}\alpha \cdot {\sin^{2}}\beta +{\cos^{2}}\alpha {\cos^{2}}\beta =1\)




知识点:第五章 三角函数


参考答案:左边 \( = {\sin ^2}\alpha (1 - {\sin ^2}\beta ) + {\sin ^2}\beta + {\cos ^2}\alpha {\cos ^2}\beta \) \(={\sin^{2}}\alpha {\cos^{2}}\beta +{\sin^{2}}\beta +{\cos^{2}}\alpha {\cos^{2}}\beta \)

\( = {\cos ^2}\beta ({\sin ^2}\alpha + {\cos ^2}\alpha ) + {\sin ^2}\beta \) \( = {\cos ^2}\beta + {\sin ^2}\beta \) \( = 1\) \( = \) 右边


解析:



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