sin(2α+π/12)=sin (α+π/6+α-π/12)
=sin(α+π/6)cos(α-π/12)+cos(α+π/6)sin(α-π/12)
cos(α+π/6)=4/5,α為銳角,則sin(α+π/6)=3/5
cos(α-π/12)=cos(α+π/6-π/4)
=cos(α+π/6)cosπ/4+sin(α+π/6)sinπ/4
=4/5*√2/2+3/5*√2/2
=7√2/10
sin(α-π/12)=sin(α+π/6-π/4)
=sin(α+π/6)cosπ/4-cos(α+π/6)sinπ/4
=3/5*√2/2-4/5*√2/2
=-√2/10
帶入原式得:
sin(2α+π/12)=sin(α+π/6)cos(α-π/12)+cos(α+π/6)sin(α-π/12)
=3/5*7√2/10+4/5*(-√2/10)
=17√2/50不好意思，我已經(jīng)解決。另外，建議你這樣做：sin（2α+π/12）=sin〔2（α+π/6）-π/4〕=√2/2（sin（2α+）π/3）-cos（2α+π/3））.然后用二倍角解決即可。很久沒做題，公式記得不多。解題方法有很多，大家互相交流交流，多謝提點(diǎn)。