Question

求下列各式的值。

（1）tan15°/（1-tan＾2 15°）；
（2）sinπ/24·cosπ/24·cosπ/12.
请写清解题过程。

Answer 1

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

Answer 2

（2）原式=1/2sinπ/12·cosπ/12=1/4sinπ/6=1/8