Ответы
Ответ дал:
1
B4)
Тогда
![-5sint cost=-5*(-0.32)=1.6 -5sint cost=-5*(-0.32)=1.6](https://tex.z-dn.net/?f=-5sint+cost%3D-5%2A%28-0.32%29%3D1.6)
B5) формулы:
![1) cos(arccos a )= a; \ \ -1 \leq a \leq 1 \\ \\ 2) sin \alpha = \sqrt{1-cos^2 \alpha } ; \ \ 0 \leq \alpha \leq \pi 1) cos(arccos a )= a; \ \ -1 \leq a \leq 1 \\ \\ 2) sin \alpha = \sqrt{1-cos^2 \alpha } ; \ \ 0 \leq \alpha \leq \pi](https://tex.z-dn.net/?f=+1%29+cos%28arccos++a++%29%3D+a%3B++%5C+%5C+-1+%5Cleq++a+%5Cleq+1++%5C%5C++%5C%5C+2%29+sin+%5Calpha+%3D+%5Csqrt%7B1-cos%5E2+%5Calpha+%7D+%3B+%5C+%5C+0+%5Cleq++%5Calpha++%5Cleq++%5Cpi+)
Решение:
![\sqrt{15} sin(arccos \frac{1}{4} )= \sqrt{15}* \sqrt{1-cos^2(arccos \frac{1}{4} )} = \\ \\ = \sqrt{15} * \sqrt{1-( \frac{1}{4})^2 } = \sqrt{15}* \sqrt{1- \frac{1}{16} } = \sqrt{15}* \sqrt{ \frac{15}{16} } = \sqrt{15}* \frac{ \sqrt{15}}{4} =\\ \\ = \frac{15}{4} =3.75 \sqrt{15} sin(arccos \frac{1}{4} )= \sqrt{15}* \sqrt{1-cos^2(arccos \frac{1}{4} )} = \\ \\ = \sqrt{15} * \sqrt{1-( \frac{1}{4})^2 } = \sqrt{15}* \sqrt{1- \frac{1}{16} } = \sqrt{15}* \sqrt{ \frac{15}{16} } = \sqrt{15}* \frac{ \sqrt{15}}{4} =\\ \\ = \frac{15}{4} =3.75](https://tex.z-dn.net/?f=+%5Csqrt%7B15%7D+sin%28arccos+%5Cfrac%7B1%7D%7B4%7D+%29%3D+%5Csqrt%7B15%7D%2A+%5Csqrt%7B1-cos%5E2%28arccos+%5Cfrac%7B1%7D%7B4%7D+%29%7D+%3D+%5C%5C+%5C%5C+%3D++%5Csqrt%7B15%7D+%2A++%5Csqrt%7B1-%28+%5Cfrac%7B1%7D%7B4%7D%29%5E2+%7D+%3D+%5Csqrt%7B15%7D%2A++%5Csqrt%7B1-++%5Cfrac%7B1%7D%7B16%7D++%7D+%3D+%5Csqrt%7B15%7D%2A+%5Csqrt%7B+%5Cfrac%7B15%7D%7B16%7D+%7D+%3D+%5Csqrt%7B15%7D%2A+%5Cfrac%7B+%5Csqrt%7B15%7D%7D%7B4%7D+%3D%5C%5C+%5C%5C+%3D+%5Cfrac%7B15%7D%7B4%7D+%3D3.75)
Тогда
B5) формулы:
Решение:
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