『急求』1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4)1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4) 需要详细过程.sinxcosx+cosx^2-1/2=1/2sin2x+1/2cos2x=√2/2sin(2x+π/4)希望能把sinxcosx+cosx^2-1/2怎么等于1/2sin2x+1/2cos2x也一
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![『急求』1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4)1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4) 需要详细过程.sinxcosx+cosx^2-1/2=1/2sin2x+1/2cos2x=√2/2sin(2x+π/4)希望能把sinxcosx+cosx^2-1/2怎么等于1/2sin2x+1/2cos2x也一](/uploads/image/z/9308148-60-8.jpg?t=%E3%80%8E%E6%80%A5%E6%B1%82%E3%80%8F1%2F2sin2x%2B1%2F2cos2x%E6%80%8E%E4%B9%88%E7%AD%89%E4%BA%8E%E2%88%9A2%2F2sin%282x%2B%CF%80%2F4%291%2F2sin2x%2B1%2F2cos2x%E6%80%8E%E4%B9%88%E7%AD%89%E4%BA%8E%E2%88%9A2%2F2sin%282x%2B%CF%80%2F4%29+%E9%9C%80%E8%A6%81%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B.sinxcosx%2Bcosx%5E2-1%2F2%3D1%2F2sin2x%2B1%2F2cos2x%3D%E2%88%9A2%2F2sin%282x%2B%CF%80%2F4%29%E5%B8%8C%E6%9C%9B%E8%83%BD%E6%8A%8Asinxcosx%2Bcosx%5E2-1%2F2%E6%80%8E%E4%B9%88%E7%AD%89%E4%BA%8E1%2F2sin2x%2B1%2F2cos2x%E4%B9%9F%E4%B8%80)
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『急求』1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4)1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4) 需要详细过程.sinxcosx+cosx^2-1/2=1/2sin2x+1/2cos2x=√2/2sin(2x+π/4)希望能把sinxcosx+cosx^2-1/2怎么等于1/2sin2x+1/2cos2x也一
『急求』1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4)
1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4) 需要详细过程.
sinxcosx+cosx^2-1/2
=1/2sin2x+1/2cos2x
=√2/2sin(2x+π/4)
希望能把sinxcosx+cosx^2-1/2怎么等于1/2sin2x+1/2cos2x也一起写上
『急求』1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4)1/2sin2x+1/2cos2x怎么等于√2/2sin(2x+π/4) 需要详细过程.sinxcosx+cosx^2-1/2=1/2sin2x+1/2cos2x=√2/2sin(2x+π/4)希望能把sinxcosx+cosx^2-1/2怎么等于1/2sin2x+1/2cos2x也一
=√2/2(√2/2cos2x+√2/2sin2x)
=√2/2(sinπ/4cosπ/4+sinπ/4cosπ/4)
=√2/2sin(2x+π/4)
((1+cos2α)/(3sin2α))*((2sin^2(α))/cos2α)的化简 急
求证:cos2αcos2β=1/2{cos2(α+β)+cos2(α-β)}
(SIN2X)^2+SIN2X+COS2X=1求角X
(SIN2X)^2+SIN2X+COS2X=1求角X
sin^2(x-π/4)=[1-cos2(x-π/4)]/2=(1+sin2x)/2
1+2cos²Ø-cos2Ø【急,过程】
sin²α*sin²β+cos²α*cos²β-1/2cos2αcos2β=(1-cos2α)/2*(1-cos2β)/2+(1+cos2α)/2*(1+cos2β)/2-1/2cos2α*cos2β=1/4(1+cos2α*cos2β-cos2α-cos2β)+1/4(1+cos2α*cos2β+cos2α+cos2β)-1/2cos2α*cos2β=1/4+1/
1.2sin2x-sinxcosx-cosx2=12.7cosx+3cos2x=03.sin2x/cos2x=cos2x/sinx4.(1+sinx)/(1+cosx)=1/2不好意思,第一题写错了,应该2sin2x+-sinxcosx-cos2=1
fx=(sinx-cos)sin2x/sinx 急1求 定义域及最大值 2求 递增区间
数学二倍角的正弦,余弦,正切公式y=2分之1*sin2x+sinx的平方,求值域.f(x)=cos(x+3分之2派)+2*cos2分之x的平方求值域
y=1-(sin2x)^2求周期,
已知向量a=(1+sin2x,sinx-cosx),b=(1,sinx+cosx),设函数f(x)=ab,若f(θ)=8/5,求cos2(п/4-2θ)
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sin2x=8/5,求cos2(π/4-2x)额 不做了 抓狂 学得好烂
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sin2α+2cos2α=-1 求tanα
下列函数中周期为2的奇函数是 1,y=sin2x 2,y=cos2πx 3,y=cos2(πx-π/4)-1/2 4,y=tanπ/2x先到先得