解:(1)原式=·sin(-sin α)
=·(-sin α)
=·(-cos α)(-sin α)
=-cos2α.
(2)原式=sin(-α-π)cos+cos α·
cos[-(2π-α)]
=sin[-(α+π)]cos+cos αcos(2π-α)
=-sin(α+π)sin α+cos αcos α
=sin2α+cos2α
=1.
利用诱导公式证明恒等式 [典例] 求证:
=.
[证明] 左边=
=
==.