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Question Number 125288 by Mathgreat last updated on 09/Dec/20

Commented by Mathgreat last updated on 09/Dec/20

$$\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}$$

Commented by mr W last updated on 09/Dec/20

$$youcannotprove,because{it}^{\prime }snot$$$$true.$$$$justput\beta =0,youwillget$$$${\cos }^2\alpha =1$$$$butthisisnottrue.$$

Answered by mathmax by abdo last updated on 10/Dec/20

$${\cos }^2(\mathrm{x}+\mathrm{y})+{\cos }^2(\mathrm{x}-\mathrm{y})-{\cos }^2\mathrm{x}.{\cos }^2\mathrm{y}$$$$={(\mathrm{cosx}\mathrm{cosy}-\mathrm{sinxsiny})}^2+{(\mathrm{cosxcosy}+\mathrm{sinx}\mathrm{siny})}^2-{\cos }^2\mathrm{x}{\cos }^2\mathrm{y}$$$$={\cos }^2\mathrm{x}{\cos }^2\mathrm{y}-2\mathrm{cosxcosy}\mathrm{sinxsiny}+{\sin }^2{\mathrm{xsin}}^2\mathrm{y}+{\cos }^2{\mathrm{xcos}}^2\mathrm{y}$$$$+2\mathrm{cosx}\mathrm{cosysinx}\mathrm{siny}+{\sin }^2{\mathrm{xsin}}^2\mathrm{y}-{\cos }^2{\mathrm{xcos}}^2\mathrm{y}$$$$={\sin }^2\mathrm{x}{\sin }^2\mathrm{y}+{\cos }^2\mathrm{x}{\cos }^2\mathrm{y}\left(\mathrm{not}\mathrm{equal}\mathrm{to}1\right)$$

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