Given the vector a = (COSA, Sina), vector b = (2, - 1), if vector a is perpendicular to vector B, find (SiNx cosx) / (SiNx + cosx) If a is perpendicular to B, find (SiNx cosx) / (SiNx + cosx) 2. If | A-B | = 2, X belongs to (0, π / 2), find the value of sin (x + π / 4)

Given the vector a = (COSA, Sina), vector b = (2, - 1), if vector a is perpendicular to vector B, find (SiNx cosx) / (SiNx + cosx) If a is perpendicular to B, find (SiNx cosx) / (SiNx + cosx) 2. If | A-B | = 2, X belongs to (0, π / 2), find the value of sin (x + π / 4)

1. ∵ vector a ⊥ vector B, ᙽ 2cosx SiNx = 0,  TaNx = 2,  (SiNx cosx) / (SiNx + cosx) = (tanx-1) / (TaNx + 1) = (2-1) / (2 + 1) = 1 / 3.2.

How to get the cosx = 0 of cosx (COSA + sin β) + SiNx (COS β - Sina) + root 2

Cosx (COSA + sin β) + SiNx (COS β - Sina) + √ 2cosx = 0cosx (COSA + sin β + √ 2) + SiNx (COS β - Sina) = 0. This is an identity with X as variable and a, β as constant. Then the coefficient of each variable should be 0  cosa + sin β + √ 2 = 0 and cos β - Sina = 0 ﹥ cosa + sin β = - √ 2 and cos β - si

The square reduction of root [(1-sina * SiNx) - - (COSA * cosx)

√[(1-sinasinx)²-(cosacosx)²]=√(1-2sinasinx+sin²asin²x-cos²acos²x)=√[1-2sinasinx+(1-cos²a)sin²x-cos²a(1-sin²x)]=√(sin²a+cos²a-2sinasinx+s...

Given the function f (x) = SiNx / 2 cosx / 2 + cos? X / 2 - 2, find the maximum and minimum value of FX on [π, 17 π / 25] Why does every step of the detailed process need The interval is [π, 17 π / 12]

From the angle doubling formula, f (x) = 0.5sinx + (1 + cosx) / 2-2
=0.5(sinx+cosx)-1.5
=0.5 √ 2Sin (x + π / 4) - 1.5 (by sum difference formula)
Wrong interval?
According to the above formula, the maximum value and the minimum value can be obtained

The function y = SiNx |sinx|+|cosx| cosx+tanx |The range of tanx|is______ .

From the meaning of the question, we need to discuss the quadrant where the angle is, and determine the symbol,
When the angle X is in the first quadrant, y = 1 + 1 + 1 = 3,
When the angle is in the second image limit, y = 1-1-1 = - 1,
When the angle is in the third quadrant, y = - 1-1 + 1 = - 1,
When the angle is in the fourth quadrant, y = - 1 + 1 - 1 = - 1
So the answer is: {- 1, 3}

The functions y = g (x) and f (x) = loga (x + 1) (0

Because of the symmetry about the origin, G (- x) + F (x) = 0g (- x) = - loga (x + 1), so GX = - loga (1-x) FX + Gx = loga (1 + x) - loga (1-x) first of all, the FX domain of the function can be determined to be that x is greater than - 1 and less than 1fx = loga (1 + x) - loga (1-x) f (- x) = loga (1-x) - loga (1 + x) FX + F

The graphs of the functions y = g (x) and f (x) = loga (x + 1) (a > 1) are symmetric about the origin (1) Write the analytic expression of y = g (x); (2) If the function f (x) = f (x) + G (x) + m is an odd function, try to determine the value of real number M; (3) When x ∈ [0,1), there is always f (x) + G (x) ≥ n. find the value range of real number n

(1) Let m (x, y) be any point on the graph of function y = g (x), then the symmetric point of M (x, y) about the origin is n (- x, - y) n. on the image of function f (x) = loga (x + 1) (2) ∵ f (x) = loga (x + 1) - loga (1-x) + m is an odd function

The image of the given function f (x) = loga (1-mx / x-1), (a > 0, and a ≠ 1) is symmetric about the origin When a ＞ 1, X ∈ (R, A-2), the value range of F (x) is (1, + ∞)

f(-x)=-f(x)
loga[(1+mx)/(-x-1)]=-loga[(1-mx)/(x-1)]
(1+mx)/(-x-1)=(x-1)/(1-mx)
m²x²-x²=0
m=±1
When x = 1, (1-mx) / (x-1) = - 1 does not hold
∴m=-1
f(x)=loga[(x+1)/(x-1)]
(a-2+1)/(a-2-1)=a
a²-4a+1=0
a=2±√3
∵a>1
∴a=2+√3
r-1=0
∴r=1

Let f (x) = loga (x + 1) (a > 1) and the images of F (x) and G (x) are symmetric about the origin. (1) solve the inequality 2F (x) + G (x) > = o (2) If f (x) + G (x) > = m holds when (1) holds, find the value range of M

(1) G (x) = - f (- x), x + 1 > 0 and - x + 1 > 0 = > - 12F (x) + G (x) ≥ o, that is, 2loga (x + 1) - loga (- x + 1) ≥ 0, that is, loga [(x + 1) ^ 2 / (- x + 1)] ≥ 0
∵ a > 1, ᙽ 2 / (- x + 1) ≥ 1, ∵ - 10, ᙽ (x + 1) ^ 2 ≥ - x + 1 = > x ≤ - 3 or ≥ 0, and - 1 (2) f (x) + G (x) = loga (x + 1) - loga (- x + 1) = loga [(x + 1) / (- x + 1)] ≥ M
When x ∈ [0,1), the minimum value of H (x) = (x + 1) / (- x + 1) = (x-1 + 2) / (- x + 1) = - 1-2 / (x-1) ∈ [1, + ∞), and a > 1, / / F (x) + G (x) ∈ [0, + ∞), m ≤ [f (x) + G (x)] = 0, that is, m ≤ 0

It is known that f (x) = loga [(1-mx) / X-1] (a > 0A ≠ 1) is an odd function, and the monotonicity of F (x) on (1, ∞) is judged It is known that f (x) = loga [(1-mx) / X-1] (a > 0A ≠ 1) is an odd function. The monotonicity of F (x) on (1, ∞) is determined by finding the value of M

f(-x)
=loga[(1+mx)/-x-1]
=-f(x)
=loga[(x-1)/(1-mx)]
1-m^2x^2=1-x^2
(1-m^2)x^2=0
m=±1.
When m = 1, the real number = - 10 and is a minus function
Loga t on R +
When 0