Q&A on sin α μ+cos α max

• Maximum

  According to Cauchy inequality, there are sinα+sinβ≤√[2((sinα)^2+(sinβ)^2)]=√2cosγ， Similarly, sinβ+sinγ≤√2cosγ， sinγ+sinα≤√2cosβ. So 2 (sin α+sin β+sin γ) ≤ √ 2 (cos α+cos β+cos γ) So (sin α+sin β+sin γ)

  mathematics

  2 answers

• sinθ＝？ When cos θ cos θ sin θ takes the maximum value

  Sint>0 and cost>0 should be limited, and the mean inequality of three variables should be used. (costcostsint)^2=(cost)^2*(cost)^2*(sint)^2 =1/2[(cost)^2*(cost)^2*2(sint)^2] =<1/2*{[(cost)^2+(cost)^2+2(sint)^2]/3}^...

  mathematics

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• Known sin α+sin β=(√ 2)

  u^2=(cosα+cosβ)^2; (1) 1/2=(sinα+sinβ)^2. (2) (1)+(2) u^2+1/2=2+2cos(α-β) u^2=3/2+2cos(α-β)≤3/2+3=7/2  u=the maximum value of cos α+cos β √ (7/2)

  mathematics

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• Find the maximum value of cos α (sin α+cos α) 0 ＜α ＜ 90

  cosα（sinα+cosα） =cosαsinα+cos^α =1/2*sin(2α)+1/2(1+cos2α） =1/2+1/2(sin(2α)+cos(2α）) =1/2+(√2)/2sin(2α+45°) 0＜α＜90° 0＜2α＜180° 45°＜2α+45°＜135° (√2)/2

  mathematics

  1 answer

• Maximum

  Let y=f (x)=(sinx-1)/(cosx-2) sinx-ycosx=1-2y, so according to Cauchy inequality, [1 ^ 2+(- y) ^ 2] * [(sinx) ^ 2+(cosx) ^ 2]>=[1 * sinx+(- y) * cosx] ^ 2 1+y ^ 2>=(1-2y) ^ 2 3y ^ 2-4y=<0 y (3y-4)=<0=

  mathematics

  2 answers

• How to find the maximum and minimum of cos ^ 2 (a) * sin (a)?

  Let y=cos ^ 2a * sina Then y=sina - (sina) ^ 3 Derivative of y=cosa-3cosa (sina) ^ 2 Let cosa-3cosa (sina) ^ 2=0, then cosa=0 or sina=± √ 3/3 When cosa=0, y=0, When sina=√ 3/3, y=2 √ 3/9 When sina=- √ 3/3, y=2 √ 3/

  mathematics

  1 answer

• Find the maximum value of function f (θ)=(sin θ - 1)/(cos θ - 2)

  Let y=(sint-1)/(cost-2) --->ycost-2y=sint-1 --->sint-ycost=1-2y Divide both sides by √ (y ^ 2+1) to get sin(t-f)=(1-2y)/√(y^2+1)(tanf=y) So | 1-2y |/√ (y ^ 2+1)=<1 --->(1-2y)^2=3y^2-4y...

  mathematics

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• Maximize function

  y=f(x)=(sinx-1)/(cosx-2) --->sinx-1=ycosx-2y --->sinx-ycosx=1-2y --->√(1+y^)*sin(x-T)=(1-2y)...... Where: tanT=y ---> sin(x-T)=(1-2y)/√(1+y^) ....∈ [-1,1] -...

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  4 answers

• High School Mathematics

  cosαcosβsin(α+β) =[cos（α+β）+cos（α-β）]/2≤[cos（α+β）+1]/2 So cos α cos β sin (α+β) ≤ sin (α+β) [cos (α+β)+1]/2 The purpose of such scaling is to unify, that is, to transform the original formula into a function of α+β Let α+β=t, t ∈ (0,1) Then take the derivative, and simply calculate s

  mathematics

  2 answers

• Find the maximum and minimum of function y=cos (2arcsinx)+sin [arcsin (2x+1)]

  It's not necessary to ask the same question twice at the same time.

  mathematics

  2 answers

• Let the complex number z=cos θ - sin θ+√ 2+i (cos θ+sin θ)

  Solution: |z|＾=（cosx-sinx+√2）＾+（cosx+sinx）＾ =（cosx）＾+（sinx）＾-2sinxcosx+2+2√2（cosx+sinx） +（cosx）＾+（sinx）＾+2sinxcosx =4+2√2（cosx+sinx） =4+4[（√2/2）cosx+（√2/2）si...

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• Senior One Mathematics (Function Problem)

  Let sina+cosa=t, then sinacosa=(t ^ 2-1)/2, where the range of t is negative root 2<=t<=positive root 2 y=sinαcosα+sinα+cosα =(t^2-1)/2+t=(t^2)/2+t-1/2 =(t+1) ^ 2/2-1 because negative radical 2<=t<=positive radical 2 So when t=positive root sign 2, y gets the maximum value 1/

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  8 answers

• Maximum

  If 15 ° - x/4=α, then f(x)＝sin(30°+α)sin(30°-α)cosα ＝cosα[1/4·(cosα)^2-3/4·(sinα)^2] ≤1/4， ∴f(x)|max＝1/4。

  mathematics

  1 answer

• Seeking solutions

  The answer of zhang3141592 is correct Or coordinates of point B (Xb, Yb) meet (Xb+2)^2+(Yb+2)^2=1 So the locus of point B is a circle with O (- 2, - 2) as the center and radius r as 1 The maximum value of AB is | AO |+r =√(5^2+4^2) +1=1+√41

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  3 answers

• mathematics

  Find the maximum value of function y=(sin ^ 2x+1) (cos ^ 2x+3) To be simple, sin ^ 2x+1 should be maximum, cos ^ 2x+3 should be maximum ∨ The maximum value of | sinx | is 1 ∨ The maximum value of sin ^ 2x+1 is 1+1=2 |The maximum value of cosx | is 1 ∨ The maximum value of cos ^ 2x+3 is 1+3=4 That is, y=(sin ^ 2x+1) (cos ^ 2x+3

  mathematics

  2 answers

• Given the vector a (cos θ, sin θ), b (√ 2+sin θ, - cos θ), where θ∈ R, 1), find | a+b

  Given the vector a (cos θ, sin θ), b (√ 2+sin θ, - cos θ) where θ∈ R, (1) find the maximum value of | a+b | (2) If | a+b |=√ 5, And θ∈ (- π/4, π/4) to find cos (θ+π/4) Solution: Vector a+vector b=(√ 2+cos θ+sin θ, sin θ - cos θ) |Vector a+vector b | t=(√ 2+cos

  mathematics

  1 answer

• Senior two in mathematics!

  f（x）＝（sinθ－１）/（cosθ－２） Represents the slope of the line connecting point P (cos θ, sin θ) and point (1,2) on the unit circle Draw two tangents One is Y=1, and the slope is 0 According to the nature of the relationship between two tangents The tangent value of half of their included angle is 1/2 Then the tangent value of their included angle is 2 * (1/2)/[1 - (1/2) * (1/2)]=4/3 ====>f（x...

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• Linear equation/12

  Using the universal formula in the triangle, sin θ=2t/(1+t ^ 2), cos θ=(1-t ^ 2)/(1+t ^ 2) Let y=F (θ), Then y=[2t/(1+t ^ 2) - 1]/[(1-t ^ 2)/(1+t ^ 2) - 2] =(t^2-2t+1)/(3t^2+1) (3y-1)t^2+2t+(y-1)=0 △=4-4(3y-1)(y-...

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• The problem of finding the maximum

  y=(sinT-1)/(cosT-2) --->sinT-ycosT=1-2y --->sin(T-A)=(1-2y)/√(1+y^2) (tanA=y) |sin(T-A)|=<1 --->|1-2y|/√(1+y^2}=<1 --->(1-2y)^2=<1+y^2 --->3y^2-4y=<0 ...

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  2 answers

• Senior two math problems

  y=[(sinθ)^2] *cosθ y^2=[(sinθ)^4] *[cosθ^2]=1/2*[(sinθ)^2] *[(sinθ)^2] *[2cosθ^2] ≤1/2*[(sinθ)^2 +(sinθ)^2 +2cosθ^2)/3]^3=1/2*8/27=4/27 y≤(2√3)/9 When and only when

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  5 answers

• trigonometric function

  |a+b|^2=(1+sinx)^2+(1+cosx)^2 =2+2(sinx+cosx) =2+2√2(sin(x+45) <=2+2√2 therefore |A+b | Maximum=√ (2+2 √ 2)

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  1 answer

• 4sin θ, then the maximum value of t is?

  t=3cosθ-4sinθ=5[(3/5)cosθ-(4/5)sinθ] =5[(cosφcosθ-sinφsinθ] =5 cos (φ+θ), so the maximum value of t is 5 Where φ=arcsin (4/5), i.e. sin φ=4/5, cos φ=3/5

  mathematics

  4 answers

• What is the maximum value of 2cos α+sin α?

  √5[sina*(1/√5）+cosa*(2/√5）] =√ 5sin (a+t), where tank=2 So the maximum value=√ 5

  mathematics

  2 answers

• Trigonometric function proof

  (cosα/1+sinα)-(sinα/1+cosα) =[(cosa/2)^2-(sina/2)^2]/(sina/2+cosa/2)^2-[2sina/2cosa/2]/[1+2(cosa/2)^2-1] =[(cosa/2)-(sina/2)]/(sina/2+cosa/2)-[sina/2]...

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• Maximum

  Let t=sin α+cos α=√ 2sin (α+π/4), then - √ 2 ≤ t ≤ √ 2, sin α * cos α=(t ² - 1)/2 Y=g (t)=0.5t ²+at-0.5, symmetry axis t=- a (1) If - a ≤ - √ 2, that is, when a ≥ √ 2, Y (min)=g (- √ 2)=0.5 - (√ 2) a, Y(max)=g(√2)=...

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