Question

Помогите, пожалуйста, задание на фото

Answer 1

B4)
$sint+cost=0.6\\ (sint+cost)^2=0.6^2 \\ sin^2t+2sint cost +cos^2t=0.36 \\ 2sint cost+1=0.36 \\ 2sint cost=0.36 -1 \\ 2sint cost=-0.64 \\ sint cost= \frac{-0.64}{2} \\ sint cost=-0.32$ 
Тогда 

$-5sint cost=-5*(-0.32)=1.6$

B5) формулы:

$1) cos(arccos a )= a; \ \ -1 \leq a \leq 1 \\ \\ 2) sin \alpha = \sqrt{1-cos^2 \alpha } ; \ \ 0 \leq \alpha \leq \pi$

Решение:

$\sqrt{15} sin(arccos \frac{1}{4} )= \sqrt{15}* \sqrt{1-cos^2(arccos \frac{1}{4} )} = \\ \\ = \sqrt{15} * \sqrt{1-( \frac{1}{4})^2 } = \sqrt{15}* \sqrt{1- \frac{1}{16} } = \sqrt{15}* \sqrt{ \frac{15}{16} } = \sqrt{15}* \frac{ \sqrt{15}}{4} =\\ \\ = \frac{15}{4} =3.75$