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

Commented by Mathgreat last updated on 09/Dec/20

Prove

$$\boldsymbol{{Prove}} \\ $$

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

you can not prove, because it′s not  true.  just put β=0, you will get  cos^2  α=1  but this is not true.

$${you}\:{can}\:{not}\:{prove},\:{because}\:{it}'{s}\:{not} \\ $$$${true}. \\ $$$${just}\:{put}\:\beta=\mathrm{0},\:{you}\:{will}\:{get} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\alpha=\mathrm{1} \\ $$$${but}\:{this}\:{is}\:{not}\:{true}. \\ $$

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

cos^2 (x+y)+cos^2 (x−y)−cos^2 x.cos^2 y  =(cosx cosy−sinxsiny)^2  +(cosxcosy+sinx siny)^2 −cos^2 x cos^2 y  =cos^2 x cos^2 y−2cosxcosy sinxsiny+sin^2 xsin^2 y+cos^2 xcos^2 y  +2cosx cosysinx siny +sin^2 xsin^2 y −cos^2 xcos^2 y  =sin^2 x sin^2 y+cos^2 x cos^2 y (not equal to 1)

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

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