Given the vector M = (SiNx, cosx), n = (COS θ, sin θ), and m · n = radical 10 / 10, if θ = π / 8, find the value of sin2x

Given the vector M = (SiNx, cosx), n = (COS θ, sin θ), and m · n = radical 10 / 10, if θ = π / 8, find the value of sin2x

It can be seen that sin (x + θ) = root 10 / 10, because the problem can use double angle, first calculate the square value of COS (2x + 2 θ) = 1-2sin (x + θ), and then expand the cosine twice to get the difference relationship between cos2x and sin2x. Then, use the relationship between the sum of squares of cos2x and sin2x to calculate the result

Given the function FX = 2 (SiNx + cosx). Cosx, then the minimum positive period of FX is

Because f (x) = 2 (SiNx + cosx). Cosx = 2sinxcosx + 2 (cosx) ^ 2 = sin2x + 2 (cosx) ^ 2-1 + 1
=sin2x+cos2x+1
So the minimum positive period of F (x) is π

The minimum positive period of the function FX = (SiNx cosx) ^ 2 The process should be detailed

f(x)=(sinx-cosx)²
=sin²x-2sinxcosx+cos²x
=1-2sinxcosx
=1-sin2x
So the minimum positive period T = 2 π / 2 = π
Answer: π

The minimum positive period of the function FX = SiNx - (cosx SiNx) is

f(x)=2sinx+cosx
=√ 5sin (x + φ) (where cos φ = 2 / √ 5, sin φ = 1 / √ 5)
The minimum positive period is t = 2 π / 1 = 2 π

The minimum positive period of the function y = root 2sinxcosx + cos ^ 2x-1 / 2

y=√2/2*sin2x+(1+cos2x)/2-1/2
=√2/2*sin2x+1/2*cos2x
=√3/2*sin(2x+z)
Where Tanz = (1 / 2) / (√ 2 / 2) = √ 2 / 2
So t = 2 π / 2 = π

The known function f (x) = sin2 (x 2+π 12）+ 3sin（x 2+π 12）cos（x 2+π 12）-1 2． (I) find the range of F (x); (II) if f (x) (x > 0), the image and straight line y = 1 The abscissa of intersection point 2 is x1, x2 The sum of the first 2n terms of the sequence {xn}

(I) f (x) = 1 − cos (x + π 6) 2 + 32sin (x + π 6) − 12 = 32sin (x + π 6) − 12cos (x + π 6) = SiNx, so the range of F (x) is [- 1,1] (II) from the symmetry and periodicity of sine curve, we can know that X1 + X22 = π 2, X3 + X42 = 2 π + π 2, X2N − 1 + x2n2 = 2 (n − 1) π + π 2 ﹤ X1 + x2 +

Let f (x) = sin ^ 2 (x / 2 + Pai / 12) + Radix 3sin (x / 2 + Pai / 12) cos (x / 2 + Pai / 12) - 1 / 2. (1) find the range of F (x). (2) if f (...) Let f (x) = sin ^ 2 (x / 2 + Pai / 12) + Radix 3sin (x / 2 + Pai / 12) cos (x / 2 + Pai / 12) - 1 / 2. (1) find the value range of F (x). (2) if the abscissa of the intersection point of the image of F (x) (x > 0) and the straight line y = 1 / 2, the abscissa from large to small is x1, X2 Xn, find the sum of the first 2n terms of the sequence {xn}

2) f=1/2=sin(x)=kπ+π/6 or 2kπ+π/6
The sum of 2n terms can be divided into odd and even n terms
Odd = A1N + n (n-1) d / 2 = π * n / 6 + π * n (n-1) = π (n ^ 2-5n / 6)
S even = A1N + n (n-1) d / 2 = 7 π * n / 6 + π * n (n-1) = π (n ^ 2 + n / 6)
S2n = s even + s odd = π (2n ^ 2-2n / 3)

What is the value range of the function y = cos ＾ 2x-2sinxcosx sin ＾ 2x

y=cos＾2x-2sinxcosx-sin＾2x
=cos2x-sin2x
=√2(sinπ/4cos2x-cosπ/4sin2x)
=√2(-2x+π/4)
So: the value range of the function y = cos ＾ 2x-2sinxcosx sin ＾ 2x is [- √ 2, √ 2]

If x ∈ [0, π], find the value range of function

If the coefficients in front of SiNx and cosx can't be directly expressed by sin or cos, then they can be expressed by (a? 2 + B? 2) under the root sign, that is, (the square of two roots and three + 2?) = 4

Y = (SiNx) ^ 2 + 4cosx + 1 y = radical (x ^ 2 + 1) + radical [(X-2) ^ 2 + 4]

1) Y = 4sinx + 1 / 2 * cosx-4 = √ (16 + 1 / 4) * sin (x + α) - 4, where Tan α = 1 / 8
Therefore, the range is: [- √ 65 / 2-4, √ 65 / 2-4]
2)y=(sinx)^2+4cosx+1=1-(cosx)^2+4cosx+1=-(cosx-2)^2+6
Because - 1