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Support001
@Support001
November 2021
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arccos(cos230)+arcsin(sin(-400)+arctg(tg110)
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oganesbagoyan
Verified answer
Arccos(cos230°) +arcsin(sin(-400°))+arctq(tq110°) =
arccos(-cos50°) +arcsin(-sin(360°+40°)) +arctq(tq180°-70°) =
=π - arccos(cos50°) - arcsin(sin40°) - arctq(tq70°) =
π - 50π/180 - 40π/180 - 70π/180 =
π/9
.
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Verified answer
Arccos(cos230°) +arcsin(sin(-400°))+arctq(tq110°) =arccos(-cos50°) +arcsin(-sin(360°+40°)) +arctq(tq180°-70°) =
=π - arccos(cos50°) - arcsin(sin40°) - arctq(tq70°) =
π - 50π/180 - 40π/180 - 70π/180 =π/9.