函數(shù)y=cos2x-sin2x+2sinx•cosx的最小正周期為 ,此函數(shù)的值域?yàn)? .
【答案】
分析:利用二倍角的余弦公式化簡(jiǎn)函數(shù)y=cos
2x-sin
2x+2sinx•cosx,再化為一個(gè)角個(gè)一個(gè)三角函數(shù)的形式,求出函數(shù)的最小正周期,和值域.
解答:解:函數(shù)y=cos
2x-sin
2x+2sinx•cosx=cos2x+sin2x=
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sin(2x+
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)
所以函數(shù)函數(shù)y=cos
2x-sin
2x+2sinx•cosx的最小正周期為:
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=π
函數(shù)的值域?yàn)椋?img src="http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131023212001895370869/SYS201310232120018953708012_DA/3.png">
故答案為:π;
點(diǎn)評(píng):本題考查三角函數(shù)的周期性及其求法,兩角和與差的正弦函數(shù),二倍角的正弦,二倍角的余弦,考查計(jì)算能力,三角函數(shù)的化簡(jiǎn),是基礎(chǔ)題.