СРочно найти значение
sin 72 *cos 12 - cos 72 * cos 78
Ответы на вопрос
Ответил mionkaf1
0
Пишите градусы, а то можно подумать, что это в радианах!!!
![sin(72^{circ})*cos(12^{circ})-cos(72^{circ})*cos(78^{circ})= \ \
=frac{1}{2}(sin(84^{circ})+sin(60^{circ}))- frac{1}{2}(cos(-6^{circ})+cos(150^{circ}))= \ \
=frac{1}{2}(sin(84^{circ})+ frac{ sqrt{3} }{2} )- frac{1}{2}(cos(6^{circ})- frac{ sqrt{3} }{2})= \ \
=frac{sin(84^{circ})}{2}+ frac{ sqrt{3} }{4}- frac{cos(60^{circ})}{2}+ frac{ sqrt{3} }{4}= frac{sin(84^{circ})+ sqrt{3}-cos(6^{circ}) }{2} \ \
sin(90^{circ}-6^{circ})=sin84^{circ} ; ;;sin84^{circ}=cos(6^{circ}); \ \
frac{cos(6^{circ})+ sqrt{3}-cos(6^{circ}) }{2}= boxed {frac{ sqrt{3} }{2}}](https://tex.z-dn.net/?f=sin%2872%5E%7Bcirc%7D%29%2Acos%2812%5E%7Bcirc%7D%29-cos%2872%5E%7Bcirc%7D%29%2Acos%2878%5E%7Bcirc%7D%29%3D++%5C++%5C+%0A%3Dfrac%7B1%7D%7B2%7D%28sin%2884%5E%7Bcirc%7D%29%2Bsin%2860%5E%7Bcirc%7D%29%29-+frac%7B1%7D%7B2%7D%28cos%28-6%5E%7Bcirc%7D%29%2Bcos%28150%5E%7Bcirc%7D%29%29%3D++%5C++%5C+%0A%3Dfrac%7B1%7D%7B2%7D%28sin%2884%5E%7Bcirc%7D%29%2B+frac%7B+sqrt%7B3%7D+%7D%7B2%7D+%29-+frac%7B1%7D%7B2%7D%28cos%286%5E%7Bcirc%7D%29-+frac%7B+sqrt%7B3%7D+%7D%7B2%7D%29%3D++%5C++%5C+%0A%3Dfrac%7Bsin%2884%5E%7Bcirc%7D%29%7D%7B2%7D%2B+frac%7B+sqrt%7B3%7D+%7D%7B4%7D-+frac%7Bcos%2860%5E%7Bcirc%7D%29%7D%7B2%7D%2B+frac%7B+sqrt%7B3%7D+%7D%7B4%7D%3D+frac%7Bsin%2884%5E%7Bcirc%7D%29%2B+sqrt%7B3%7D-cos%286%5E%7Bcirc%7D%29+%7D%7B2%7D+%5C++%5C+%0Asin%2890%5E%7Bcirc%7D-6%5E%7Bcirc%7D%29%3Dsin84%5E%7Bcirc%7D+%3B+%3B%3Bsin84%5E%7Bcirc%7D%3Dcos%286%5E%7Bcirc%7D%29%3B+%5C++%5C+%0A+frac%7Bcos%286%5E%7Bcirc%7D%29%2B+sqrt%7B3%7D-cos%286%5E%7Bcirc%7D%29+%7D%7B2%7D%3D+boxed+%7Bfrac%7B+sqrt%7B3%7D+%7D%7B2%7D%7D++ "sin(72^{circ})*cos(12^{circ})-cos(72^{circ})*cos(78^{circ})= \ \
=frac{1}{2}(sin(84^{circ})+sin(60^{circ}))- frac{1}{2}(cos(-6^{circ})+cos(150^{circ}))= \ \
=frac{1}{2}(sin(84^{circ})+ frac{ sqrt{3} }{2} )- frac{1}{2}(cos(6^{circ})- frac{ sqrt{3} }{2})= \ \
=frac{sin(84^{circ})}{2}+ frac{ sqrt{3} }{4}- frac{cos(60^{circ})}{2}+ frac{ sqrt{3} }{4}= frac{sin(84^{circ})+ sqrt{3}-cos(6^{circ}) }{2} \ \
sin(90^{circ}-6^{circ})=sin84^{circ} ; ;;sin84^{circ}=cos(6^{circ}); \ \
frac{cos(6^{circ})+ sqrt{3}-cos(6^{circ}) }{2}= boxed {frac{ sqrt{3} }{2}}")
Новые вопросы