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Ответ:

Применяем формулы понижения степени и произведения

тригонометрических функций :  

[tex]\bf cos^2x=\dfrac{1+cos2x}{2}\ \ \Rightarrow \ \ 1+cos2x=2cos^2x[/tex]   .

[tex]\bf cos^2\alpha +cos^2\beta -cos(\alpha +\beta )\cdot cos(\alpha -\beta )=\\\\\\=\dfrac{1+cos2\alpha }{2}+\dfrac{1+cos2\beta }{2}-\dfrac{1}{2}\cdot (cos2\alpha -cos2\beta )=\\\\\\=\dfrac{1}{2}\cdot (1+cos2\alpha +1+cos2\beta )-\dfrac{1}{2}\cdot (cos2\alpha -cos2\beta )=\\\\\\=\dfrac{1}{2} \cdot (2+cos2\alpha +cos2\beta -cos2\alpha+cos2\beta )=\\\\\=\dfrac{1}{2}\cdot \Big(2+2cos2\beta \Big)=1+cos2\beta =2cos^2\beta=[/tex]  

[tex]\bf =\dfrac{1}{2}\cdot (2+2cos2\beta )=1+cos2\beta =2cos^2\beta[/tex]