Cos θ + sin θ - 2 =? (maximum)

Cos θ + sin θ - 2 =? (maximum)

cosa+sina-2
=√2(√2/2cosa+√2/2sina)-2
=√2sin(a+π/4)+2
Because: - 1 ≤ sin (a + π / 4) ≤ 1
So we can get: √ 2 + 2 ≤ sin (a + π / 4) + 2 ≤√ 2 + 2
The maximum value of cosq + sina-2 is: √ 2 + 2
Note: sin π / 4 = cos π / 4 = √ 2 / 2

How to find the maximum value of sin θ + cos θ? To understand the process

Sin θ + cos θ = root 2 * [(root 2) / 2 * sin θ + (root 2) / 2 * cos θ] = root 2 * (cos45 ° sin θ + sin45 ° cos θ) = root 2 * sin (θ + 45 °) because the maximum value of sin (θ + 45 °) is 1, the maximum value of this is root 2

How to find the maximum value of sin θ + sin θ * cos θ

Derivation:
Let f (x) = sin θ + sin θ * cos θ
Then f '(x) = cos θ + (COS θ) ^ 2 - (sin θ) ^ 2
=cosθ+2(cosθ)^2-1
=2(cosθ+1/4)^2-9/8
The positive and negative conditions of F '(x) are studied,
When and only if cos θ = 1 / 2 and sin θ = √ 3 / 2, the maximum value of F (x) is 3 √ 3 / 4

Given the function y = sin Λ 4x + 2 ∫ 3sinxcosx cos∧ 4x, find the minimum value, positive period and minimum value of the function

Y = sin Λ 4x + 2 √ 3sinxcosx - cos Λ 4x (root sign before 3, not integral?) = sin ∧ 4x - cos ∧ 4x + √ 3sin2x = (sin ∧ x-cos ∧ 4x + √ 3sin2x = √ 3sin2x - cos2x = 2 * sin (2x - π / 6) period T = 2 π / w = 2 π / 2 =

The minimum positive period of the function y = cos ^ 4x sin ^ 4x + 2 is A.π B.2π C.π/2 D.π/4

y=(cos²x－sin²x)(cos²x＋sin²x)＋2
=cos²x－sin²x＋2
=cos2x＋2
The minimum positive period is 2 π / 2 = π

If the function f (x) = (radical 3) cos (3x + a) - sin (3x + a) is odd, then Tana is equal to

For odd functions, f (0) = 0
So √ 3cosa Sina = 0
sina=√3cosa
tana=sina/cosa=√3

Given that sin (π - a) = 4 / 5, a belongs to (0, π / 2), find the value of sin2a cos ^ 2 A / 2 and find the value of function f (x) = 5 / 6 Cosa sin2x-1 / 2cos2x The second question is to find the value of the function f (x) = 5 / 6 * cosa * sin2x-1 / 2 * cos2x

(1) From sin (π - a) = 4 / 5,
sina=sin(π-a)=4/5
And a belongs to (0, π / 2),
So cosa is greater than 0, and (COSA) ^ 2 + (Sina) ^ 2 = 1,
Cosa = 3 / 5,
So sin2a - cos ^ 2 A / 2 = 2sinacosa - (0.5 + COSA)
=24/25 - 0.5 -3/5
= -0.14
(2) According to cosa = 3 / 5,
f(x)=5/6* cosa *sin2x-1/2*cos2x
=1/2*sin2x -1/2*cos2x
=0.5√2*sin(2x - π/4)
What does it mean to find the value of this function? Is it to find its value range?
Obviously, the range of sin (2x - π / 4) is [- 1,1],
Therefore, the range of F (x) is [- 0.5 √ 2,0.5 √ 2]

Find the function f (x) = (sin quartic x + cos quartic x + sin? Xcos? X) / (2-2sinxcosx) - (sinxcosx) / (2) + (cos2x / 4) Minimum positive period minimum value Maximum

f(x)=（sin^4x+cos^4x+sin²xcos²x）/（2-2sinxcosx）-（sinxcosx）/（2）+（cos2x/4）
=（sin^4x+cos^4x+2sin²xcos²x-sin²xcos²x）/（2-2sinxcosx）-1/4*sin2x+1/4*cos2x
=[(sin²x+cos²x)²-sin²xcos²x）/（2-2sinxcosx）-1/4*sin2x+1/4*cos2x
=(1-sin²xcos²x）/（2-2sinxcosx）-1/4*sin2x+1/4*cos2x
=(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx)-1/4*sin2x+1/4*cos2x
=(1+sinxcosx)/2-1/4*sin2x+1/4*cos2x
=1/2+1/4*sin2x-1/4*sin2x+1/4*cos2x
=1/2+1/4*cos2x
T=2π/2=π
-1

Given SiNx cosx = 1 / 5, find the value of the following formula 1. Sin ^ 3x cos ^ 3x 2. Cosx / (1-sinx) - SiNx / (1 + cosx) I'm sorry. wrote it wrong. yes It is known that SiNx cosx = 1 / 5, 1.sin^3x-cos^3x 2.cosx/(1-sinx)-sinx/(1+cosx)

sinx-cosx=1/5
sinx ²+cosx ²-2sinx cosx =(sinx-cosx)²=1/25
sinxcosx=12/25
sin^3x-cos^3x=(sinx-cosx)(sin^2x +cos^2x+sinxcosx)=(sinx-cosx)(1+12/25)
=1/5*37/25=37/125
2)
cosx/(1-sinx)-sinx/(1+cosx)
=cosx(1+cosx)-sinx(1-sinx)/[(1-sinx)(1+cosx)]
=(cosx-sinx +1)/[1-sinx+cosx-sinxcosx]
=(-1/5+1)/(1-1/5-12/25)
=4/5 / 8/25
=4/5*25/8
=5/2

Given SiNx + cosx = 1 / 3, find sin ^ 3x + cos ^ 3x

From (SiNx + cosx) ^ 3 = (SiNx) ^ 3 + (cosx) ^ 3 + 3sinxcosx (SiNx + cosx) ^ 3 = (SiNx + cosx) ^ 3 - 3sinxcosx (SiNx + cosx) = (1 / 3) ^ 3-3 * (1 / 3) sinxcosx = 1 / 27-sinxcosx (1) and (SiNx + cosx) ^ 2 = (1