1,6a-0,3p и 2,1a-0,1p при а=-1,p=-2

Question

Алгебра: 1,6a-0,3p и 2,1a-0,1p при а=-1,p=-2​

fdnk1 · Accepted Answer

1)sin20⁰cos10⁰+cos20⁰sin10⁰=
=1/2[sin(20⁰+10⁰)+sin(20⁰-10⁰)]+1/2[sin(10⁰+20⁰)+sin(10⁰-20⁰)]=
=1/2[sin30⁰+sin10⁰]+1/2[sin30⁰+sin(-10⁰)]=sin30⁰+1/2sin10⁰-1/2sin10⁰=
=sin30⁰=1/2;
2)sinπ/5cos4π/5+cosπ/5sin4π/5=
=1/2[sin(π/5+4π/5)+sin(π/5-4π/5)]+1/2[sin(4π/5+π/5)+sin(4π/5-π/5)]=
=1/2[sinπ+sin(-3π/5)]+1/2[sinπ+sin(3π/5]=
=sinπ-1/2sin(3π/5)+1/2sin(3π/5)=
=sinπ=0;
3)cos80⁰cos10⁰+sin80⁰cos10⁰=
=1/2[cos(80⁰-10⁰)+cos(80⁰+10)⁰]+1/2[sin(80⁰+10⁰)+sin(80⁰-10⁰)]=
=1/2[cos70⁰+cos90⁰]+1/2[sin90⁰+sin70⁰]=
=1/2[cos70⁰+0]+1/2[1+sin70⁰]=1/2cos70⁰+1/2sin70⁰+1/2;
4)cos3π/8sinπ/8+cosπ/8sin3π/8=
=1/2[sin(π/8+3π/8)+sin(π/8-3π/8)]+1/2[sin(3π/8+π/8)+sin(3π/8-π/8)]=
=1/2[sin(π/2)+sin(-π/4)]+1/2[sin(π/2)+sin(π/4)]=
=sin(π/2)-1/2sin(π/4)+1/2sin(π/4)=sin(π/2)=1;

1,6a-0,3p и 2,1a-0,1p при а=-1,p=-2​

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