If the vector PQ = (4 / 3, - 2 / 3) Vector op × vector OQ=

If the vector PQ = (4 / 3, - 2 / 3) Vector op × vector OQ=

PQ = oq-op = (2cosb Sina, 2sinb COSA) = (4 / 3, - 2 / 3), so 2cosb Sina = 4 / 3, 2sinb cosa = - 2 / 3, the sum of the two formulas after the square, 4 + 1-4 (sinacosb + cosasinb) = 20 / 9, the solution of sinacosb + cosasinb = 25 / 36, so OP * OQ = 2 (sinacosb + cosasinb) = 25 / 18

Given the function f (x) = 2Sin (π - x) cos α (1), find the minimum positive period of F (x) (2) find the maximum and minimum value of F (x) on the interval [- π / 6, π / 2] As the title

The period of sin (π - x) is 2 π, the period of cos α is 2 π, the minimum positive period is 2 π, (2)

F (x) = 2Sin (π - x) sin (π / 2-x) find the minimum positive period of F (x) and find the maximum and minimum value of F (x) on the interval [- π / 6, π / 2] f(x)=2sin(π-x )sin(π/2-x ) Finding the minimum positive period of F (x) The minimum of the sum of π / F on the interval

F (x) = 2Sin (π - x) sin (π / 2-x) = 2sinxcosx = sin (2x), so the minimum positive period of F (x) t = 2 π / 2 = π f (x) = sin (2x) in the interval [- π / 6, π / 2] is equivalent to the maximum and minimum value of F (y) = siny in the interval [- π / 3, π], and f (y) = siny in the interval [- π / 3, π]

The function y = 2Sin (x + π) 12)cos(x+π 4) The maximum and minimum values of are () A. 2，-2 B. 3 2，−1 Two C. 3 2，1 Two D. 1 2，−3 Two

According to the product sum difference formula of trigonometric function:
y＝2sin(x+π
12)cos(x+π
4)=sin（x+π
12+x+π
4）+sin（x+π
12-x-π
4）
=sin（2x+π
3）-1
Two
When sin (2x + π)
3) When = 1, the function takes the maximum value, which is 1
2, when sin (2x + π)
3) When = - 1, the function takes the minimum value, which is - 3
2．
Therefore, D

Find the maximum and minimum of the function f (x) = 2Sin ^ x + cos ^ x-4sinx + 4

1： F (x) = 2Sin ^ x + cos ^ x-4sinx + 4 = 2Sin ^ 2xs + 1-sin ^ 2x-4sinx + 4 = sin ^ 2x-4sinx + 5 = (sinx-2) ^ 2 + 1 when SiNx = - 1, f (x) takes the maximum value, f (x) takes the maximum value, f (x)

The function y = 2Sin squared x + cos squared x-2sinx-1 to find the maximum and minimum value of Y and when the function gets the minimum value, the set process of X values is entrusted to you 3Q

Equi strip formula: it can be reduced to the square x-2sinx of y = sin, and then it can be changed into the square - 1 of y = (sinx-1), thus the maximum value can be obtained. When SiNx = - 1, the maximum value is 3! When SiNx = 1, the minimum value is - 1

Given that f (x) = cos (2x - π / 2) + 2Sin (x - π / 4) sin (x + π / 4), the equation of image symmetry axis is obtained It is known that f (x) = cos (2x - π / 2) + 2Sin (x - π / 4) sin (x + π / 4), Solving the equation of image symmetry axis

(x) = cos (π / 2-2x) + 2Sin (x-π / 4) cos (π / 2 - (x + π / 4)) = sin2x + 2Sin (x-π / 4) cos (π / 4-x) = sin2x + 2Sin (x-π / 4) cos (π / 4-x) = sin2x + 2Sin (x-π / 4) cos (x-π / 4) cos (x-π / 4) = sin2x + sin (2x-π / 2) = sin2x-sin (π / 2-2x) = sin2x-cos2x = sin2x-cos2x = root number 2 * sin (2x-π / π / 4 / 4) cos (π / 4-4-x / 4-x = sin2x + 2Sin 4) so the symmetry axis is x = 3 π / 8 + K π / 2

Simplify (1) cos α * Tan α (2) 1-2sin ^ 2 α 2cos ^ 2-1

(1)cosα*tanα
=cosα*sinα/cosα
=sinα
(2) 1-2sin ^ 2 α 2cos ^ 2-1
=Cos2 α in Cos2 α
=1

Simplify sin? β + cos β to the fourth power + sin? β cos? β

The original formula = sin? β + cos? β (sin? β + cos? β)
=sin²β+cos²β
=1

Given the vector M = (2, a √ 3cosx), n = (COS Λ 2x, 2sinx), function f (x) = m · n-1, find the minimum positive period and monotone increasing interval of function f (x)

F (x) = cos2x root 3sin2x, will you