利用半角公式,求下列各式的值：1.sinπ/12-cosπ/122.tanπ/8+tan3π/8

Question

利用半角公式,求下列各式的值：
1.sinπ/12 - cosπ/12
2.tanπ/8 + tan3π/8

Answer 1

1.sinπ/12 < cosπ/12
(sinπ/12 - cosπ/12 )^2=1-2sinπ/12cosπ/12
=1-sin(π/6)=1-1/2=1/2
故：sinπ/12 - cosπ/12 ＝-√2/2
2.tanπ/8 + tan3π/8=tanπ/8 +tan(π/2-π/8)
=tanπ/8+cotπ/8
=sin(π/8)/cos(π/8)+cos(π/8)/sin(π/8)通分
＝1/[sin(π/8)cos(π/8)]
=2/sin(π/4)
=2√2