Question

Помогите решить уравнения arcsin(-1/2)+arcsin1 2 arccos √3/2 - arctg √3 arccos1-arcsin(-√3/2) arccos(√2/2)+2arcsin1/2

Answer 1

$1)arcSin(-\frac{1}{2})+arcSin1=-\frac{\pi }{6} +\frac{\pi }{2}=-\frac{\pi }{6}+\frac{3\pi }{6}=\frac{2\pi }{6}=\frac{\pi }{3}\\\\2)2arcCos\frac{\sqrt{3}}{2}-arctg\sqrt{3}=2*\frac{\pi }{6}-\frac{\pi }{3}=\frac{\pi }{3}-\frac{\pi }{3}=0\\\\3)arcCos1-arcSin(-\frac{\sqrt{3}}{2}=0-(-\frac{\pi }{3} )=\frac{\pi }{3}\\\\4)arcCos(\frac{\sqrt{2}}{2}+2arcSin\frac{1}{2}=\frac{\pi }{4}+2*\frac{\pi }{6}=\frac{\pi }{4}+\frac{\pi }{3} =\frac{3\pi+4\pi}{12}=\frac{7\pi}{12}$

Answer 2

Давайте по порядку решим каждое уравнение:

1. arcsin(-1/2):

arcsin(-1/2) = -π/6

2. arcsin(1):

arcsin(1) = π/2

3. 2 * arccos(√3/2):

arccos(√3/2) = π/6

2 * arccos(√3/2) = 2 * π/6 = π/3

4. arctg(√3):

arctg(√3) = π/3

5. arccos(1):

arccos(1) = 0

6. arcsin(-√3/2):

arcsin(-√3/2) = -π/3

7. arccos(√2/2):

arccos(√2/2) = π/4

8. 2 * arcsin(1/2):

2 * arcsin(1/2) = 2 * π/6 = π/3

Теперь, объединим все значения:

arcsin(-1/2) + arcsin(1) + 2 * arccos(√3/2) - arctg(√3) + arccos(1) - arcsin(-√3/2) + arccos(√2/2) + 2 * arcsin(1/2)

= (-π/6) + (π/2) + (π/3) - (π/3) + 0 - (-π/3) + (π/4) + (π/3)

= π/2 + π/4 + (π/3 - π/3) - (π/6 - π/3)

= π/2 + π/4 - π/6

= (3π + 2π - 4π) / 12

= π / 12

Ответ: π / 12