Посмотри график построения функции в учебнике и просто подставь числа
a√3 внести под корень
1. a ≥ 0 a√3 = √a²√3 = √(3a²)
2. a < 0 a√3 = - √a²√3 = - √(3a²)
1a)1/2(sin30+sin40)=1/2*1/2+1/2sin40=1/4+1/2sin40
b)1/2cos60+1/2cos14=1/2*1/2+1/2cos14=1/4+1/2cos14
c)1/cos12-1/2cos60=1/2cos12-1/4
2a)sin(90-20)cos(90-25)-sin25cos20=sin25cos20-sin25cos20=0
b)cos(π/2-7π/18)*cos(π/2-4π/9)-sin7π/36*sin5π/36=
sin7π/18*sin4π/9-sin7π/36*sin5π/36=2sin7π/36cos7π/36sin4π/9-sin7π/36*sin5π/36=
sin7π/36(2cos7π/36sin4π/9-sin5π/36)= sin7π/36(2*1/2sinπ/4+2*1/2sin23π/36-sin5π/36)=sin7π/36(√2/2+sin23π/36-sin5π/36)=
sin7π/36(√2/2+2sinπ/2cos7π/9)=sin7π/36(√2/2+2cos7π/9)=√2/2sin7π/36+
2sin7π/36*cos7π/9=√2/2sin7π/36+2*1/2sin(-7π/12)+2*1/2sin37π/9=
√2/2sin7π/36-sin7π/12+sin(π+π/36)=√2/2sin7π/36-sin7π/12-sinπ/36
По моему решается уравнением .x+x=333-0;2x=333;x=166,5
[√(n+2)*(√(n+2)+√(n-2))]/[√(n+2)*(√(n+2)-√(n-2))] +
+[√(n+2)*(√(n+2)-√(n-2))]/[√(n+2)*(√(n+2)+√(n-2))]=
=(√(n+2)+√(n-2))/(√(n+2)-√(n-2))+(√(n+2)-√(n-2))/(√(n+2)+√(n-2))=
(n+2+2√(n²-2)+n-2+n+2-2√(n²-2)+n-2)/[(√(n+2)+√(n-2))(√(n+2)-√(n-2))]=
=4n/(n+2-n+2)=4n/4=n
√(√2-1)²+2-√2=√2-1+2-√2=1