Доказать тождество:
1) [tex]tga(1+cos2a)=sin2a
[/tex]
2) [tex] \frac{(1-2cos^2a)(2sin^2a-1)}{4sin^2a*cos^2a} =ctg^22a
[/tex]
1) [tex]tga(1+cos2a)=sin2a
[/tex]
2) [tex] \frac{(1-2cos^2a)(2sin^2a-1)}{4sin^2a*cos^2a} =ctg^22a
[/tex]
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Verified answer
Решение.................=2sinacos²a/cosa=2sinacosa=sin2a
2)(1-2cos²a)(2sin²a-1)/4sin²acos²a=(2sin²-1-4sin²acos²a+2cos²a)/4sin²acos²a=
=(2-1-4sin²acos²a)/sin²2a=(1-sin²2a)/sin²2a=cos²2a/sin²2a=cotg²2a