cos12/5兀等于多少
cos12分之5π怎么算 -
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cos2x = 1- 2(sinx)^2 (sinx)^2 = (1-cos2x)/2 sinx = ±√[(1-cos2x)/2]x=π/12 sin(π/12)= √[(1-cos(π/6))/2]=√[(1-√3/2)/2]=√(2-√3)/2 =(√3-1)/(2√2)=√2(√3-1)/4 cos(5π/12)=sin(π/12) =√2(√3-1)/4 where 2-√3 = (希望你能满意。
请采纳，
cos2x = 1- 2(sinx)^2 (sinx)^2 = (1-cos2x)/2 sinx = ±√[(1-cos2x)/2]x=π/12 sin(π/12)= √[(1-cos(π/6))/2]=√[(1-√3/2)/2]=√(2-√3)/2 =(√3-1)/(2√2)=√2(√3-1)/4 cos(5π/12)=sin(π/12) =√2(√3-1)/4 where 2-√3 = (希望你能满意。
cos12π/5=cos(2π+2π/5)=cos2π/5 cos11π/2=cos(6π-π/2)=cos（π/2)=cosπ/2=0
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计算:sec [12分之5派],等于多少? -
见下图：
这是化简题吧cos5/12派=cos432º=cos72º
cos5╱12派等于??用公式怎么算? -
【注：sin2x=2sinxcosx. sin(π-x)=sinx】解：原式=（1/2)×2sin(5π/12)cos(5π/12)=(1/2)sin(5π/6)=(1/2)sin[π-(π/6)]=(1/2)sin(π/6)=(1/2)×(1/2)=1/4.∴原式=1/4。
cos（5π/12）cos（π/4+π/6）cosπ/4cosπ/6-sinπ/4sinπ/6 =√2/2*√3/2-√2/2*1/2 =√6/4-√2/4 =（√6-√2）4
cos12兀/5得多少?拜托拜托!!谢谢! -
cos12兀它具有周期性，因为周期等于2所以cos12兀=cos2=0，所以结果等于0
cos5/12兀=cos75° =cos(30°+45°)再用和的余弦公式计算可得到答案。