化简[sin(x+nπ)+sin(x-nπ)]/[sin(x+nπ)*cos(x－nπ)]

Question

化简[sin(x+nπ)+sin(x-nπ)]/[sin(x+nπ)*cos(x－nπ)]

Answer 1

利用"和差化积”公式sin α+sin β=2sin[(α+β)/2]·cos[(α-β)/2]
"积化和差”公式sinαcosβ= 1/2[sin(α+β)+sin(α-β)]
得
[sin(x+nπ)+sin(x-nπ)]/[sin(x+nπ)*cos(x－nπ)]
＝[2sinx·cos(nπ)]/(1/2)[sin(2x)+sin(2nπ)]
＝＋－4sinx/sin(2x） (这个式子中cos(nπ)=+ -1,sin(2nπ)=0)
＝＋－4sinx/(2sinx·cosx)
＝＋－2/cosx
＝＋－2secx
其中的"＋－"是指正负号