cos(χ-π/4)=(cosχ+sinχ)/√2=-1/2, ∴cosχ+sinχ=-1/√2， 平方得1+sin2χ=1/2,sin2χ=-1/2, (cosχ-sinχ)^2=1-sin2χ=3/2, ∴cosχ-sinχ=土√6/2， ∴cos(5π/4-χ)+cos²(π+χ) ...
由于,-3/2π<α<-π/2则3/4π〈π/4-α〈7/4π，又因为COS(π/4-α)=3/5 〉0， 则=〉3/2π〈4/7π/4-α〈7/4π 。故SIN(π/4-α)〈0 。SIN(π/4-α)=-4/5 则SIN(π/2-2α）=2SIN(π/4-α)COS(π/4-α)=-12/25 ...
解： ∵α∈(π/4，3π/4)， ∴ -π/2＜π/4-α＜0 cos(π/4-α)=3/5 cos（α-π/4）=cos（π/4-α）=3/5 sin（π/4-α）=-4/5 sin（α-π/4）=-sin（π/4-α）=4/5 ∵β∈(0，π/4)， ...
cos(π/4+θ)cos(π/4-θ) =(1/2)[cos(π/2)+cos2θ] =(1/2)cos2θ =（√3）/4， ∴cos2θ=(√3）/2. θ∈（3π/4，π）, ∴2θ∈（3π/2，2π）， ∴2θ=11π/6， θ=11π/12， ∴sinθ+cosθ=（√2）sin(θ+π...
[(sinα)^2+(cosα)^2]^3=(sinα)^6+3(sinα)^4(cosα)^2+3(sinα)^2(cosα)^4+(cosα)^6 所以分子就是:3(sinα)^4(cosα)^2+3(sinα)^2(cosα)^4=3[(sinα)^2(cosα)^2] 分母:2[(sinα)...
sin（α+３π/４）=sin（α+（π-３π/４））= sin（α-π/４ = ５/１３ cos（π/４-β) = cos（-（β-π/４)）= cos（β-π/４)= 4/5 cos(α-β) = cos（(α-π/４）-（β-π/４)）=.............
问老师
cos（π/4-θ）cos（π/4+θ）=1/8 cos（π/4-θ）sin（π/4-θ）=1/8 2*1/2cos（π/4-θ）sin（π/4-θ）=1/8 sin(π/2-2θ)=1/4 sinπ/2cos2θ-cosπ/2sin2θ=1/4 cos2θ=1/4 cos4θ=2cos^22θ-1...
提示：2（cosθ-sinθ）=2\2\sin(Π/4-θ)=2\2\sin[(Π/4-θ/2)-θ/2] =2\2\sin(Π/4-θ/2)cosθ/2-cos(Π/4-θ/2)sinθ/2; 1+sinθ+cosθ=2(cosθ/2)^2+2sinθ/2cosθ/2=2cosθ/2(cosθ/2...
1-(sinα)^6-(cosα)^6 =1-sin^6α-cos^6α = sin^2α(1-sin^4α) + cos^2α(1-cos^4α) = sin^2α[sin^2α(1-sin^2α) + cos^2α] + cos^2α[cos^2α(1-cos^2α)+sin^2α] =sin^...
由已知可,α在45-90度之间.cosα为正.cos(派/4-α)=4/5=> cosα+sinα=4/5*根号2 => sin α-cosα=3/5*根号2 cosα=根号2/10 cos(派/4-α)=4/5 =>sin(派/4-α)=-3/5 sin(3派/4+β)=5/13 =>cos...
sin[(4n-1)π/4－α］+cos[(4n+1)π/4－α] =sin[nπ－α-π/4］+cos[nπ－α+π/4] =sin(nπ－α)cosπ/4-cos(nπ－α)sinπ/4+con(nπ－α)cosπ/4-sin(nπ－α)sinπ/4 =0
π0 因此tanα=√(41²-9²)/9=40/9 所以tan(π/4-α)=(1-40/9)/(1+40/9)=-31/49.
α为第二象限角，且cosα=－3/5 sinα >0 ==> sinα =4/5 ==> sin(α＋л/3)=sinα*cosл/3+cosα*sinл/3=(4-4*根号3)/10 sin(3л/4-α)=sin3л/4*cosα-cos3л/4*sinα =(根号2)/10
因为,0＜β＜π/4，π/4＜α＜3π/4， 所以,-π/2<π/4-α<0,3π/4<3π/4+β<π 又因为cos(π/4-α）=3/5 ,sin (3π/4+β）=5/13 所以,sin(π/4-α）=-4/5,cos(3π/4+β）=-12/13 所以, cos(α-β） =cos(β-α）...
若3π/4＜θ＜5π/4化简 √{cosπ/4*sin(3π/4-θ)[sin(π-θ)-sin(θ-π/2)]} 化简───────----───────────── sin(θ+π/4) 解答如下，单击放大：
z=cosθ+[5/4-(sinθ)^2]i (0≤θ<2π)，求|z|、argz的范围 x=cosθ,y=5/4-(sinθ)^2=1/4+(cosθ)^2 |z|=根号（x^2+y^2）=根号（（(cosθ)^2+3/4）^2-1/2） 0≤θ<2π，0≤(cosθ)^2≤1,1/4≤|z|≤根...
θ∈(0, π/4), π/4+θ∈(π/4, π/2), sin(π/4+θ )=12/13 cosθ=cos[(π/4+θ)-π/4]=cos(π/4+θ)cosπ/4+sin(π/4+θ)sinπ/4 =5/13*√2/2+12/13*√2/2=17√2/26 sin(π/4-θ)=cos[...
解：原题cos(α/2)应为cot(α/2) y={[1-cos(π-2α)]/[cot(α/2)-tan(α/2)]}-[cos(π/4-β)]^2 =[1+cos2α]/[cos(α/2)/sin(α/2)-sin(α/2)/cos(α/2)]-[cos(π/4-β)]^2 =2（cosα）＾×...
sin(α π/2)=cosa=-√5/5 所α∈(π/2π). sina=2√5/5 [cos2(π/4 α/2)-cos2(π/4-α/2)]/[sin(π-α) cos(3π α)]=[(1 cos(π/2 α))/2-(1 cos(π/2-α))/2]/[sina-cosa]=-sina/...
lz你没学过和差化积公式吗？ sin α+sin β=2sin[(α+β)/2]·cos[(α-β)/2] sin α-sin β=2cos[(α+β)/2]·sin[(α-β)/2] cos α+cos β=2cos[(α+β)/2]·cos[(α-β)/2] cos α-cos β=-2sin[...
M∩N≠Φ即下面方程组有解： {cosα=cosβ {λ+sinβ=4-(cosα)^2 ∴λ=4-(cosβ)^2-sinβ =(sinβ-1/2)^2+11/4 ∈[11/4，5]。
可以看出(3π/4+β)-(π/4-α)=π/2+α+β 所以想到求cos（π/2+α+β)=-sin（α+β） 0＜β＜π/4所以3π/4<3π/4+β<π cos(3π/4+β）=-12/15 π/4＜α＜3π/4所以-π/2<π/4-α<0 sin(π/4-α）=-4/5 cos(π/4-α）...
解：π/4＜α+β＜π sin(α+β)＞0. -π/2＜π/4-α＜0 3π/4＜3π/4+β＜π 所以sin(π/4-α)=-4/5 cos(3π/4+β)=-12/13 `sin[(π/4-α)-(3π/4+β)] =sin[-π/2-(α+β)] =cos(α+β) 又sin[(π/4-α)...
令z＝x+yi (x、y∈R),则 {x＝cosθ,y＝5/4-(sinθ)^2} →y＝x^2+1/4. 其中|x|≤1,画草图易知 |z|∈[1/4,√41/4]. 过原点作抛物线切线y＝kx, 代入抛物线方程整理得： x^2-kx+1/4＝0. △＝0→k＝±1, 对应的切点横坐标为1/2∈[...
∵ (3π/4+β)-(π/4-α)=π/2+(α+β), ∴ sin(α+β)=cos[(3π/4+β)-(π/4-α)]=cos(3π/4+β)cos(π/4-α)+sin(3π/4+β)sin(π/-α).而 -π/2<π/4-α<0, 3π/4<3π/4+β<π, ∴sin(π/4-α)=-...
解： tanα=2tan（α/2）/[1-tan＾（α/2）]=4/3 ∵α,β∈(0,π), ∴cosα=1/√[1+tan＾α]=3/5 sinα=4/5 ∵sin(α-β)=5/13, ∴sinαcosβ-cosαsinβ=5/13 令cosx 4x/5-5/13=（3/5）√...
解: ∵(k∈Z） ∴sin[(4k-1)*π/4—⊙]+cos(4k+1)*π/4-⊙) =sin[kπ-⊙-π/4]+cos[kπ-⊙+π/4] 令kπ-⊙=β 则:sin[kπ-⊙-π/4]+cos[kπ-⊙+π/4] =sin(β-π/4)+cos(β+π/4) =sin(β-π/4)+c...
我的解答如下：
解 α∈(π/4,3π/4),π/4-α∈(-π/2,0), ∴ sin(π/4-α)=-4/5.β∈(0,π/4),3π/4+β∈(3π/4,π),cos(3π/4+β)=-12/13,sin(α+β)=-cos[(3π/4+β)-(π/4-α)]=sin(3π/4+β)sin(π/4-α)-c...
