题目内容
已知cos(
+
)=-
,且
是第四象限角,计算:
(1)sin(2
-
);
(2)
(n∈Z).
(1)
(2)-4
解析:
∵cos(
+
)=-
,∴-cos
=-
,cos
=
,
又∵
是第四象限角,∴sin
=-
.
(1)sin(2
-
)=sin[2
+(-
)]
=sin(-
)=-sin
=
.
(2)![]()
=![]()
=![]()
=
=
=
=-4.
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题目内容
已知cos(
+
)=-
,且
是第四象限角,计算:
(1)sin(2
-
);
(2)
(n∈Z).
(1)
(2)-4
∵cos(
+
)=-
,∴-cos
=-
,cos
=
,
又∵
是第四象限角,∴sin
=-
.
(1)sin(2
-
)=sin[2
+(-
)]
=sin(-
)=-sin
=
.
(2)![]()
=![]()
=![]()
=
=
=
=-4.