\(1-2sin^2x+1-2sin^2y+2sin\left(x+y\right)=2\)
\(\Rightarrow sin^2x+sin^2y-sin\left(x+y\right)=0\)
\(\Rightarrow sin^2x+sin^2y-sinx.cosy-siny.cosx=0\)
\(\Rightarrow sinx\left(sinx-cosy\right)+siny\left(siny-cosx\right)=0\)
\(\Rightarrow\frac{sinx}{sinx+cosy}\left(sin^2x-cos^2y\right)+\frac{siny}{siny+cosx}\left(sin^2y-cos^2x\right)=0\)
\(\Rightarrow\frac{sinx}{sinx+cosy}\left(sin^2x+sin^2y-1\right)+\frac{siny}{sinx+cosy}\left(sin^2x+sin^2y-1\right)=0\)
\(\Rightarrow\left(sin^2x+sin^2y-1\right)\left(\frac{sinx}{sinx+cosy}+\frac{siny}{sinx+cosy}\right)=0\)
\(\Rightarrow sin^2x+sin^2y=1\) (ngoặc phía sau luôn dương với \(x;y\in\left(0;\frac{\pi}{2}\right)\)
\(\Rightarrow sin^2x=1-sin^2y=cos^2y\)
\(\Rightarrow sinx=cosy=sin\left(\frac{\pi}{2}-y\right)\)
\(\Rightarrow x=\frac{\pi}{2}-y\Rightarrow x+y=\frac{\pi}{2}\)
\(P=\frac{sin^4x}{y}+\frac{cos^4y}{x}=\frac{sin^4x}{y}+\frac{sin^4x}{x}=sin^4x\left(\frac{1}{x}+\frac{1}{y}\right)\)
Ủa hàm này làm gì có min nhỉ, bạn coi lại đề có nhầm ở đâu ko?
