Given the square of X + X-1 = 0, find the value of X's cubic power = 2x's quadratic power + 3

Given the square of X + X-1 = 0, find the value of X's cubic power = 2x's quadratic power + 3

x^2+x-1=0
x^2+x=1
x^3+2x^2+3
=x^3+x^2+x^2+x-x+3
=x(x^2+x)+(x^2+x)-x+3
=x+1-x+3
=4
^Represents the index

Given sin (x + π / 6) = - 3 / 5, find the value of sin 2 (π / 3-x) - sin (5 π / 6-x) It's 31 / 25

sin(x+π/6)=-3/5
Sin is an odd function
So sin (- X - π / 6) = 3 / 5
sin(π/3-x)=sin(-x-π/6+π/2)=cos(-x-π/6)=√(1-9/25)=4/5
sin(5π/6-x)=sin(-x-π/6+π)=-sin(-x-π/6)=-3/5
(see Quadrant for odd and even invariant symbols in legend)
Bring in
16/25+15/25=31/25

Given sin (x + π / 6) = 1 / 3, find the value of sin (7 π / 6 + x) + cos 2 (5 π / 6-x) It's due tomorrow

sin(7π/6+x)
=sin(π+π/6+x)
=-sin(π/6+x)
=-1/3
cos²(5π/6-x)
=1-sin²(5π/6-x)
=1-sin²[π-(5π/6-x)]
=1-sin²(π/6+x)
=8/9
So the original formula = 5 / 9

Given the = 1 / 4 of sin (x + π / 6), find the square value of sin (7 π / 6 + x) + cos (5 π / 6 - x) The answer is 11 / 16

The = 1 / 4sin (7 π / 6 + x) + [cos (5 π / 6-x)] ^ 2 = sin (π + π / 6 + x) + [cos (5 π / 6-x)] ^ 2 = sin (π + π / 6 + x) + [cos [π (6 + 6 + x)] ^ 2 = - sin (x + π / 6) + [cos (x + π / 6)] ^ 2 = - sin (x + π / 6) + {[1 - [sin (x + π / 6)] ^ 2} = - 1 / 4 + [1 - (1 / 4) ^ 2] = - 1 / 4 + [1 (1 / 4) ^ 2] = - 1 / 1 / 4 + [1 (1 / 4) ^ 2] = - 1 / 1 / 1 / 1 / 4 ^ 2] = - 1 / 4 + 1-1 / 16 = 11 / 16

If sin (x + π / 6) = 1 / 4, then the value of sin (7 π + x) + sin ^ 2 (5 π / 6-x) is given I only know that the first one is equal to 1 / 4, but I don't know how to calculate the latter one~ Sin (7 π / 6 + x) Sorry for the wrong number --

-sin(π+x)=sinx -sin(-π+x)=sinx
The front item is - 1 / 4
According to SiNx = - sin (- x), sin (5 π / 6-x) = - sin (X-5 π / 6) = - sin (x + π / 6 - π) = sin (x + π / 6) = 1 / 4
The original formula = - 1 / 4 + 1 / 16 = - 3 / 16

sin[(π/6)-x]= -3/5,π/6

Method 1: first, calculate SiNx and cosx respectively, and then calculate cos2x
From sin [(π / 6) - x] = - 3 / 5 and the formula sin (a-b) = sinacosb sinbcosa
Sin (π / 6) cosx sinxcos (π / 6) = - 3 / 5, namely cosx / 2 - (√ 3 / 2) SiNx = - 3 / 5 (1)
By π / 6

It is known that f (x) = sin (2x + π / 6) + 3 / 2, X belongs to R (1) Find the minimum positive period and monotone increasing interval of function f (x) (2) How can the image of function f (x) be transformed from the image of function y = sin2x (x belongs to R)?

1) The minimum positive period T = 2 π / 2 = π, 2K π - π / 2

It is known that f (x) = sin (2x - π / 3) - cos (2x + π / 6) ① The f (x) image is shifted to the right by M units to get the image passing through the origin and find the minimum value of M ② If x0 ∈ [- π / 12,5 π / 12] f (x0) = - 8 / 5, find the value of F (x0 + π / 4)

A:
f(x)=sin(2x-π/3)-cos(2x+π/6)
=cos(π/3)sin2x-sin(π/3)cos2x-cos(π/6)cos2x+sin(π/6)sin2x
=sin2x-√3cos2x
=2*[(1/2)sin2x-(√3/2)cos2x]
=2sin(2x-π/3)
1）
The image is shifted to the right by M units to get: F (x) = 2Sin (2x-2m - π / 3)
Through the origin: F (0) = 2Sin (- 2m - π / 3) = 0
So: 2m + π / 3 = k π
So: M = k π / 2 - π / 6 > 0
Therefore, when k = 1, the minimum value of M is π / 3
2）
f(x)=2sin(2x-π/3)=-8/5
So: sin (2x - π / 3) = - 4 / 5
So: sin (2x + π / 6 - π / 2) = - sin [π / 2 - (2x + π / 6)] = - cos (2x + π / 6) = - 4 / 5
Therefore, cos (2x + π / 6) = 4 / 5
Because:
-π/12<=x<=5π/12
-π/6<=2x<=5π/12
0<=2x+π/6<=π
So:
0<=2x+π/6<=π/2
Combined with the identity (COSA) ^ 2 + (Sina) ^ 2 = 1, the following results are obtained
sin(2x+π/6)=3/5
So:
f(x+π/4)
=2sin(2x+π/2-π/3)
=2sin(2x+π/6)
=2*(3/5)
=6/5

Given the function f (x) = sin (2x + Π / 6) + 3 / 2, X ∈ R (1) Finding the minimum positive period and monotone increasing interval of function f (x) (2) How can the image of function f (x) be transformed from the image of function y = sin2x (x ∈ R)

(1) The minimum positive period is [2 μ g / T] = w, so the period T is equal to W. (W is the coefficient before x) the monotone increasing interval of function f (x) = asin is [2K Wu - (Wu / 2), 2K Wu + (Wu / 2)] so the monotone increasing interval of function f (x) requires (2x + Π / 6) in the interval [2K Wu - (Wu / 2), 2K Wu + (Wu / 2)]

It is known that f (x) = - 1 / 2 + sin (π / 6-2x) + cos (2x - π / 3) + cos squared X (1) find the minimum positive period of F (x); (2) find the maximum value of F (x) in the interval [π / 8,5 π / 2], and find the value of X when f (x) takes the maximum value

The following is the solution: (1) f (x) = - 1 / 2 + sin (π / 6-2x) + cos (2x-π / 3) + (cosx) ^ 2 = - 1 / 2 + sin π / 6 cos2x-sin2x-sin2xcos π / 6 + cos2xcos π / 3 + sin2xsin π / 3 + sin2xsin π / 3 + (cosx) ^ 2 = (1 / 2) cos2x - (√ 3 / 2) sin2x + (1 / 2) cos2x + (3 / 2) sin2x + (3 / 2) sin2x + (2 (cos2x) [(2 (cosx) ^ 2-1]] (2 (cosx) ^ 2-1)] 1-1-1-1-1-1 / 2 (cosx) ^ 2/ 2 = cos2x + (cos2x) / 2 = (3 / 2