
求下列各式的值。
(1)tan15°/(1-tan^215°);(2)sinπ/24·cosπ/24·cosπ/12.请写清解题过程。...
(1)tan15°/(1-tan^2 15°);
(2)sinπ/24·cosπ/24·cosπ/12.
请写清解题过程。 展开
(2)sinπ/24·cosπ/24·cosπ/12.
请写清解题过程。 展开
展开全部
tan2α=(2tanα)/(1-tan^2α)
sin2α=2sinαcosα
解:
(1)tan15°/(1-tan^2 15°);
=1/2×tan30°
=1/2×√3/3
=√3/6
(2)sinπ/24·cosπ/24·cosπ/12.
=1/2×sinπ/12·cosπ/12
=1/2×1/2×sinπ/6
=1/4×1/2
=1/8
sin2α=2sinαcosα
解:
(1)tan15°/(1-tan^2 15°);
=1/2×tan30°
=1/2×√3/3
=√3/6
(2)sinπ/24·cosπ/24·cosπ/12.
=1/2×sinπ/12·cosπ/12
=1/2×1/2×sinπ/6
=1/4×1/2
=1/8
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询