Решение:
1)
sin²α + cos²α = 1
sin²α = 1 - cos²α = 1 - (-5/13)² = 1 - 25/169 = 144/169
Так как π/2 < α < π, то sin α > 0, тогда
sin α = √(144/169) = 12/13.
2)
tg α = sin α/cos α = 12/13 : (-5/13) = - (12•13)/(13•5) = - 12/5 = - 2 2/5.
3)
ctg α = 1/tg α = 1 : (-12/5) = - 5/12.
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Решение:
1)
sin²α + cos²α = 1
sin²α = 1 - cos²α = 1 - (-5/13)² = 1 - 25/169 = 144/169
Так как π/2 < α < π, то sin α > 0, тогда
sin α = √(144/169) = 12/13.
2)
tg α = sin α/cos α = 12/13 : (-5/13) = - (12•13)/(13•5) = - 12/5 = - 2 2/5.
3)
ctg α = 1/tg α = 1 : (-12/5) = - 5/12.