Question

化簡(jiǎn)：

sin2(α+π)•cos(π+α)•cos(-α-2π)

tan(π+α)•sin3( π 2 +α)•sin(-α-2π)

=______．

Answer 1

根據(jù)誘導(dǎo)公式及正弦余弦函數(shù)的奇偶性化簡(jiǎn)得：

sin2(α+π)•cos(π+α)•cos(-α-2π)

tan(π+α)•sin3( π 2 +α)•sin(-α-2π)

=

(-sinα)2•(-cosα)•cosα

tanα•cos3α•(-sinα)

=

sin2α•cos2α

sinα cosα •cos3α•sinα

=1
故答案為1．