解:(1)因為x∈(
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,
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),
所以x-
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∈(
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),
sin(x-
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)=
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=
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.
sinx=sin[(x-
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)+
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]
=sin(x-
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)cos
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+cos(x-
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)sin
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=
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×
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+
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×
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=
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.
(2)因為x∈(
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,
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),
故cosx=-
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=-
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=-
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.
sin2x=2sinxcosx=-
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,
cos2x=2cos
2x-1=-
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.
所以sin(2x+
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)=sin2xcos
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+cos2xsin
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=-
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.
分析:(1)利用x的范圍確定x-
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的范圍,進而利用同角三角函數的基本關系求得sin(x-
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)的值,進而根據sinx=sin[(x-
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)+
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]利用兩角和公式求得答案
(2)利用x的范圍和(1)中sinx的值,利用同角三角函數的基本關系求得cosx的值,進而根據二倍角公式求得sin2x和cos2x的值,
最后代入正弦的兩角和公式求得答案.
點評:本題主要考查了兩角和公式的化簡求值和同角三角函數基本關系的應用.考查了學生基礎知識的掌握和基本運算能力.