求值
解:
∵sin(π/4+2α)sin(π/4-2α)
=sin(π/4+2α)cos(π/4+2α)
=(1/2)sin(π/2+4α)
=1/4,
∴cos4α=1。
而α∈(π/4,π/2),故α=5π/12。
∴2(sinα)^2+tanα-cotα-1
=-(cos2α+2cot2α)
=-[cos(5π/6)+2cot(5π/6)]
=(5√3)/2。
由已知等式可得:
[(√2/2)(cos2α+sin2α)]×[(√2/2)(cos2α-sin2α)]=1/4
===>2cos²(2α)=1+2sin²(2α)
===>cos²(2α)=3/4
∵(π/4)<α<(π/2)
∴(π/2)<(2α)<(π)
∴cos(2α)=-√3/2
又由cos²(2α)=3/4得:(1-2sin²α)²=3/4
===>16sin²αcos²α=1
===>sinαcosα=1/4(负数的舍去)
所求等式化为:(2sin²α-1)+(sin²α-cos²α)/(sinαc
osα)
=-cos(2α)+(1-2cos²α)/(sincosα)
=-cos(2α)+4(1-2cos²α)----------代入sinαcosα=1/4得到
=-cos(2α)-4(2cos²α-1)
=-cos(2α)-4cos(2α)
=-5cos(2α)
=(5√3)/2--------------------代入cos(2α)=-√3/2得到。
。
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