1楼:匿名用户
tana=-tan(180°-a)
tanπ/5=-tan4π/5
tan2π/5=-tan3π/5
结果为0
tana=2tanπ/5则cos(a-3π/10)/sin(a-π/5)等于多少?
2楼:暖暖炊烟袅袅
结果是3。
公式一:设 α为任意角,终边相同的角的同一三角函数的值相等:
公式二:设 α为任意角, π+ α与 α的三角函数值之间的关系:
公式三:任意角- α与 α的三角函数值之间的关系:
公式四:π- α与 α的三角函数值之间的关系:
公式五: 2π-α与 α的三角函数值之间的关系:
若tanx=2tanπ\5,则cos(x-3\10π)\sin(x-π\5)=
3楼:匿名用户
tanx=2tanπ
/5cos(x-3π/10)/sin(x-π/5)
={cosxcos3π/10+sinxsin3π/10) / (sinxcosπ/5-cosxsinπ/5)
(分子分母同除以cosx)
={cos3π/10+tanxsin3π/10) / (tanxcosπ/5-sinπ/5)
={cos3π/10+2tanπ/5sin3π/10) / (2tanπ/5cosπ/5-sinπ/5)
(分子分母同乘以cosπ/5)
={cosπ/5cos3π/10+2sinπ/5sin3π/10) / (2sinπ/5cosπ/5-cosπ/5sinπ/5)
={cosπ/5cos3π/10+sinπ/5sin3π/10+sinπ/5sin3π/10) / sinπ/5cosπ/5
={cos(3π/10-π/5)+sinπ/5sin3π/10) / sinπ/5cosπ/5
={cosπ/10+sinπ/5sin3π/10) / sinπ/5cosπ/5
={cosπ/10-1/2[cos(π/5+3π/10)-cos(π/5-3π/10)]/ sinπ/5cosπ/5
={cosπ/10-1/2[cosπ/2-cos(π/10)]/ sinπ/5cosπ/5
= [(3/2)cosπ/10] /[1/2sin2π/5]
= [3cosπ/10] /[sin2π/5]
= [3cosπ/10] /[cos(π/2-2π/5)]
= [3cosπ/10]/[cosπ/10]= 3
4楼:汪汉祺
本题主要考查对 三角函数的诱导公式 考点的理解。解题过程如下:
tanx=2tanπ/5
cos(x-3π/10)/sin(x-π/5)
={cosxcos3π/10+sinxsin3π/10) / (sinxcosπ/5-cosxsinπ/5)
(分子分母同除以cosx)
={cos3π/10+tanxsin3π/10) / (tanxcosπ/5-sinπ/5)
={cos3π/10+2tanπ/5sin3π/10) / (2tanπ/5cosπ/5-sinπ/5)
(分子分母同乘以cosπ/5)
={cosπ/5cos3π/10+2sinπ/5sin3π/10) / (2sinπ/5cosπ/5-cosπ/5sinπ/5)
={cosπ/5cos3π/10+sinπ/5sin3π/10+sinπ/5sin3π/10) / sinπ/5cosπ/5
={cos(3π/10-π/5)+sinπ/5sin3π/10) / sinπ/5cosπ/5
={cosπ/10+sinπ/5sin3π/10) / sinπ/5cosπ/5
={cosπ/10-1/2[cos(π/5+3π/10)-cos(π/5-3π/10)]/ sinπ/5cosπ/5
={cosπ/10-1/2[cosπ/2-cos(π/10)]/ sinπ/5cosπ/5
= [(3/2)cosπ/10] /[1/2sin2π/5]
= [3cosπ/10] /[sin2π/5]
= [3cosπ/10] /[cos(π/2-2π/5)]
= [3cosπ/10]/[cosπ/10]
= 3诱导公式:
规律:奇变偶不变,符号看象限。即形如(2k+1)90°±α,则函数名称变为余名函数,正弦变余弦,余弦变正弦,正切变余切,余切变正切。形如2k×90°±α,则函数名称不变。
tana=2tanπ/5则cos(a-3π/10)/sin(a-π/5)=几?高中题目,求详细过程
5楼:暖暖炊烟袅袅
结果是3。
公式一:
设 α为任意角,终边相同的角的同一三角函数的值相等:
公式二:设 α为任意角, π+ α与 α的三角函数值之间的关系:
公式三:任意角- α与 α的三角函数值之间的关系:
公式四:π- α与 α的三角函数值之间的关系:
公式五: 2π-α与 α的三角函数值之间的关系:
tanα=2tan5/π 求cos(α-3/10π)/sin(α-5/π
6楼:匿名用户
tanα=2tanπ
/5cos(α-3π/10)/sin(α-π/5)=(cosαcos3π/10+sinαsin3π/10) / (sinαcosπ/5-cosαsinπ/5)
=(cos3π/10+tanαsin3π/10) / (tanαcosπ/5-sinπ/5)
=(cos3π/10+2tanπ/5sin3π/10) / (2tanπ/5cosπ/5-sinπ/5)
= / (2tanπ/5cosπ/5-sinπ/5)= / (2tanπ/5cosπ/5-sinπ/5)= / (2sinπ/5-sinπ/5)
=(3sinπ/5)/(sinπ/5)=3
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