Предмет: Алгебра,
автор: 181900000
помогите пожалуйста, четные
Приложения:
Ответы
Автор ответа:
0
423.
2)
![cos^2 alpha tg^2(- alpha )-1=cos^2 alpha *tg^2 alpha -1= \ \
=cos^2 alpha * frac{sin^2 alpha }{cos^2 alpha }-1=sin^2 alpha -1=-(1-sin^2 alpha )=-cos^ 2alpha](https://tex.z-dn.net/?f=cos%5E2+alpha+tg%5E2%28-+alpha+%29-1%3Dcos%5E2+alpha+%2Atg%5E2+alpha+-1%3D+%5C++%5C+%0A%3Dcos%5E2+alpha+%2A+frac%7Bsin%5E2+alpha+%7D%7Bcos%5E2+alpha+%7D-1%3Dsin%5E2+alpha+-1%3D-%281-sin%5E2+alpha+%29%3D-cos%5E+2alpha++ "cos^2 alpha tg^2(- alpha )-1=cos^2 alpha *tg^2 alpha -1= \ \
=cos^2 alpha * frac{sin^2 alpha }{cos^2 alpha }-1=sin^2 alpha -1=-(1-sin^2 alpha )=-cos^ 2alpha")
4)
![frac{1-tg(-x)}{sinx+cos(-x)}= frac{1+tgx}{sinx+cosx}= frac{1+ frac{sinx}{cosx} }{sinx+cosx}= \ \
= frac{cosx+sinx}{cosx(sinx+cosx)}= frac{1}{cosx}](https://tex.z-dn.net/?f=+frac%7B1-tg%28-x%29%7D%7Bsinx%2Bcos%28-x%29%7D%3D+frac%7B1%2Btgx%7D%7Bsinx%2Bcosx%7D%3D+frac%7B1%2B+frac%7Bsinx%7D%7Bcosx%7D+%7D%7Bsinx%2Bcosx%7D%3D+%5C++%5C+%0A%3D+frac%7Bcosx%2Bsinx%7D%7Bcosx%28sinx%2Bcosx%29%7D%3D+frac%7B1%7D%7Bcosx%7D+++++ "frac{1-tg(-x)}{sinx+cos(-x)}= frac{1+tgx}{sinx+cosx}= frac{1+ frac{sinx}{cosx} }{sinx+cosx}= \ \
= frac{cosx+sinx}{cosx(sinx+cosx)}= frac{1}{cosx}")
6)
![tg(-u)ctgu+sin^2u=-tgu*ctgu+sin^2u=-1+sin^2u= \ \
=-(1-sin^2u)=-cos^2u](https://tex.z-dn.net/?f=tg%28-u%29ctgu%2Bsin%5E2u%3D-tgu%2Actgu%2Bsin%5E2u%3D-1%2Bsin%5E2u%3D+%5C++%5C+%0A%3D-%281-sin%5E2u%29%3D-cos%5E2u "tg(-u)ctgu+sin^2u=-tgu*ctgu+sin^2u=-1+sin^2u= \ \
=-(1-sin^2u)=-cos^2u")
8)
![frac{tg(-x)+1}{1-ctgx}= frac{-tgx+1}{1- frac{cosx}{sinx} }= frac{1- frac{sinx}{cosx} }{ frac{sinx-cosx}{sinx} }= \ \
= frac{cosx-sinx}{cosx}* frac{sinx}{-(cosx-sinx)}=-tgx](https://tex.z-dn.net/?f=+frac%7Btg%28-x%29%2B1%7D%7B1-ctgx%7D%3D+frac%7B-tgx%2B1%7D%7B1-+frac%7Bcosx%7D%7Bsinx%7D+%7D%3D+frac%7B1-+frac%7Bsinx%7D%7Bcosx%7D+%7D%7B+frac%7Bsinx-cosx%7D%7Bsinx%7D+%7D%3D+%5C++%5C+%0A%3D++frac%7Bcosx-sinx%7D%7Bcosx%7D%2A+frac%7Bsinx%7D%7B-%28cosx-sinx%29%7D%3D-tgx++ "frac{tg(-x)+1}{1-ctgx}= frac{-tgx+1}{1- frac{cosx}{sinx} }= frac{1- frac{sinx}{cosx} }{ frac{sinx-cosx}{sinx} }= \ \
= frac{cosx-sinx}{cosx}* frac{sinx}{-(cosx-sinx)}=-tgx")
2)
4)
6)
8)
Интересные вопросы
Предмет: Другие предметы,
автор: arsten4444
Предмет: Математика,
автор: ilasisenov9
Предмет: Русский язык,
автор: gftcfrdcf2
Предмет: Физика,
автор: Caroline24
Предмет: Биология,
автор: mechtaev17