1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α)
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1.已知α是第二象限角,f(α)=[sin(3π/2-α)·cos(2π-α)·tan(π-α)]/-cos(-π-α).
(1)化简f(α)
(2)若sin(π/2 -α)=-½,求f(α)的值.
2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)
3.已知0<α<(π/4),0<β<(π/4),且3sinβ=sin(2α+β),4tan(α/2)=1-tan²(α/2),求α+β的值.
我数学很不好,遇见这种题没有思路,
(1)化简f(α)
(2)若sin(π/2 -α)=-½,求f(α)的值.
2.若π/2<α<π,化简√(1/2+1/2√1/2+1/2cos2α)
3.已知0<α<(π/4),0<β<(π/4),且3sinβ=sin(2α+β),4tan(α/2)=1-tan²(α/2),求α+β的值.
我数学很不好,遇见这种题没有思路,
1.(1)f(α)=【-cosα·(-cosα)(-tanα)】/(-cosα)=sinα,看π/2的倍数,奇变偶不变,再判断符号
(2)cosα=-1/2,∴α=2π/3,f(α)=√3/2
2.sin(α/2)利用cos2x=1-2(sinx)^2=2(cosx)^2-1逐步化简就行了
3.tanα=1/2,∴sinα=√5/5,cosα=2√5/5,化简求解即得
(2)cosα=-1/2,∴α=2π/3,f(α)=√3/2
2.sin(α/2)利用cos2x=1-2(sinx)^2=2(cosx)^2-1逐步化简就行了
3.tanα=1/2,∴sinα=√5/5,cosα=2√5/5,化简求解即得
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