1)cosa-cosb / cosa+cosb 2)sina+sinb / sina-sinb - вопрос №7264517123262 от sucker046 06.10.2021 02:31

Question

Алгебра: 1)cosa-cosb / cosa+cosb 2)sina+sinb / sina-sinb 3)sin2x+sin6x / cos2x-cos6x 4) cosa-cosb / sina+sinb 5) cos5x-cosx / sin5x+sinx 6) cos2x-cos3x / sin2x+sin3x

sucker046 · Answer

1)cosa-cosb / cosa+cosb = (-2sin((a-b)/2)*sin((a+b)/2)) / (2cos((a-b)/2)*cos((a+b)/2)) = -tg((a-b)/2)*tgn((a+b)/2)

2)sina+sinB / sina-sinB = (2sin((a+b)/2)*cos((a-b)/2)) / (2sin((a-b)/2)*cos((a+b)/2)) = tg((a+b)/2)*ctg((a-b)/2)

3)sin2x+sin6x / cos2x-cos6x = (2sin((2x+6x)/2)*cos((2x-6x)/2)) / (-2sin((2x-6x)/2)*sin((2x+6x)/2)) = -(sin4x*cos(-2x)) /(sin(-2x)*sin4x)= -ctg(-2x) = ctg2x

4) cosa-cosB / sina+sinB = (-2sin((a-b)/2)*sin((a+b)/2)) / (2sin((a+b)/2)*cos((a-b)/2)) = -tg((a-b)/2)

5) cos5x-cosx / sin5x+sinx = (-2sin((5x-x)/2)*sin((5x+x)/2)) / (2sin((5x+x)/2)*cos((5x-x)/2)) = - (sin2x*sin3x) / (sin3x*cos2x) = -tg2x

6)cos2x-cos3x / sin2x+sin3x = (-2sin((2x-3x)/2)*sin((2x+3x)/2)) / (2sin((2x+3x)/2)*cos((2x-3x)/2)) = -(sin(-1x/2)*sin(5x/2)) / (sin(5x/2)*cos(-1x/2)) = - tg(-x/2)