Question

$3( \frac{7}{9} - \frac{2}{3} ) - | - 2|$
$arcsin( \sqrt{2} - 1) + arcsin1 - \sqrt{2}$
$sin {}^{2} \frac{\pi}{5} + sin {}^{2} \frac{3\pi}{10}$
$1 - 2sin {}^{2} 15$
$(5 + \frac{3}{5} ) \div \frac{7}{10} - \frac{1}{2}$
$\sin840$
​помагите срочно

Answer 1

Пошаговое объяснение:

$1)~3(\frac{7}{9}-\frac{2}{3})-|-2|=3(\frac{7}{9}-\frac{6}{9})-2=3\cdot \frac{1}{9}-2=\frac{1}{3}-2=-(2-\frac{1}{3})=\\\\=-(1\frac{3}{3}-\frac{1}{3})=-1\frac{2}{3}\\\\2)~ \arcsin(\sqrt{2}-1)+\arcsin(1-\sqrt{2})=\arcsin(\sqrt{2}-1)+\arcsin(-(\sqrt{2}-1))=\\\\=\arcsin(\sqrt{2}-1)-\arcsin(\sqrt{2}-1)=0 \\ \\ 3)~\sin^2{\frac{\pi}{5} }+\sin^2{\frac{3\pi}{10} }=\sin^2{\frac{\pi}{5} }+\sin^2{(\frac{5\pi}{10}-\frac{2\pi}{10} )}=\sin^2{\frac{\pi}{5} }+\sin^2{(\frac{\pi}{2}-\frac{\pi}{5} )}=$

$=\sin^2{\frac{\pi}{5} }+\cos^2{\frac{\pi}{5} }=1 \\ \\ 4)~ 1-2\sin^2{15^{\circ}}=\sin^2{15^{\circ}}+\cos^2{15^{\circ}}-2\sin^2{15^{\circ}}=\cos^2{15^{\circ}}-\sin^2{15^{\circ}}=\\ \\=\cos{(2\cdot 15^{\circ})}=\cos{30^{\circ}}=\frac{\sqrt{3} }{2} \\ \\ 5)~ (5+\frac{3}{5}):\frac{7}{10}-\frac{1}{2}=(5+0,6):0,7-0,5=5,6:0,7-0,5=\\ \\=56:7-0,5=8-0,5=7,5 \\ \\ 6)~\sin{840^{\circ}}=\sin(720^{\circ}+90^{\circ}+30^{\circ})=\sin(90^{\circ}+30^{\circ})=\cos30^{\circ}=\frac{\sqrt{3} }{2}$