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PolatAlemdar
@PolatAlemdar
October 2021
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m11m
2 cos²150° - 3sin(-90°)-5ctg135°=2*(√3/2)²-3*(-1)-5*(-1)=2*(3/4)+3+5=9.5
cos150°=cos(90°+60°)=-sin60°=
-√3
2
sin(-90°)=-sin90°=-1
ctg135°=
cos135°
=
cos(90°+45°)
=
-sin45°
= -tg45°=-1
sin135° sin(90°+45°) cos45°
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Answers & Comments
cos150°=cos(90°+60°)=-sin60°=-√3
2
sin(-90°)=-sin90°=-1
ctg135°=cos135°=cos(90°+45°)=-sin45°= -tg45°=-1
sin135° sin(90°+45°) cos45°