题目内容
(Ⅰ)求值:sin690°•sin150°+cos930°•cos(-870°)+tan120°•tan1050°;(Ⅱ)已知角α的终边上有一点P(1,2),求

【答案】分析:(Ⅰ)利用诱导公式把要求的式子化为-tan30°sin30°+cos150°cos150°+tan60°tan30°,再利用特殊角的三角函数值求得结果.
(Ⅱ)根据已知角α的终边上有一点P(1,2),可得tanα=2,再根据
=
,运算求得结果.
解答:解:(Ⅰ)sin690°•sin150°+cos930°•cos(-870°)+tan120°•tan1050°
=sin(720°-30°)sin30°+cos(3×360°-150°)cos(720°+150°)+(-tan60°)tan(3×360°-30°)
=-tan30°sin30°+cos150°cos150°+tan60°tan30°=-
+
+1=
.
(Ⅱ)∵已知角α的终边上有一点P(1,2),∴tanα=2,故
=
=
=
.
点评:本题主要考查同角三角函数的基本关系的应用,诱导公式的应用,属于中档题.
(Ⅱ)根据已知角α的终边上有一点P(1,2),可得tanα=2,再根据


解答:解:(Ⅰ)sin690°•sin150°+cos930°•cos(-870°)+tan120°•tan1050°
=sin(720°-30°)sin30°+cos(3×360°-150°)cos(720°+150°)+(-tan60°)tan(3×360°-30°)
=-tan30°sin30°+cos150°cos150°+tan60°tan30°=-



(Ⅱ)∵已知角α的终边上有一点P(1,2),∴tanα=2,故




点评:本题主要考查同角三角函数的基本关系的应用,诱导公式的应用,属于中档题.

练习册系列答案
相关题目