The function y = sin (2x + π) 6）+cos（2x+π 3) The minimum positive period and maximum value of are () A. π， Two B. π，1 C. 2π， Two D. 2π，1

The function y = sin (2x + π) 6）+cos（2x+π 3) The minimum positive period and maximum value of are () A. π， Two B. π，1 C. 2π， Two D. 2π，1

∵y=sin（2x+π
6）+cos（2x+π
3）=
Three
2sin2x+1
2cos2x+（1
2cos2x-
Three
2sin2x）=cos2x，
The minimum positive period T = 2 π
2=π，ymax=1．
Therefore, B

The minimum positive period and maximum of sin (2x + π / 6) - cos (2x + π / 3)

sin(2x+∏/6)-cos(2x+∏/3)
=sin(2x+∏/6)-cos(2x+∏/2-∏/6)
=sin(2x+∏/6)+sin(2x-∏/6)
=2sin2xcos ∏/6
=√3sin2x
The maximum value is √ 3
The period is Π

The function y = sin (2x + π) 6）+cos（2x+π 3) The minimum positive period and maximum value of are () A. π， Two B. π，1 C. 2π， Two D. 2π，1

∵y=sin（2x+π
6）+cos（2x+π
3）=
Three
2sin2x+1
2cos2x+（1
2cos2x-
Three
2sin2x）=cos2x，
The minimum positive period T = 2 π
2=π，ymax=1．
Therefore, B

Cos ^ 2x / sin (π / 4 + x) cos (x + π / 4), find ① f (5 π / 12)

f(x)=cos^2x/sin( π/4+x)cos(x+π/4),
f(5π/12)=cos²(5π/6)/[(sin2π/3)cos(2π/3)]
=(3/4)/(-√3/2*1/2)
=-√3

If x = π / 12, then the value of sin ^ 2x cos ^ 2x is given

sin^2x-cos^2x=-(cos ^2 x-sin^2 x)=-cos(2x)
And substituting x = π / 12
sin^2x-cos^2x=-cos π/6=-√3/2

If cos (x + π) 6)＝−5 Then sin (π) The value of 6 − 2x) is______ .

∵2(x+π
6)+(π
6−2x)＝π
2，
∴sin(π
6−2x)＝cos2(x+π
6)＝2cos2(x+π
6)−1＝−119
169．
So the answer is: − 119
One hundred and sixty-nine

It is known that sin (x-pai / 6) = 3 / 5 (0 Math homework help users 2016-12-12 report Use this app to check the operation efficiently and accurately!

sin(x-π/6)=3/5 (0

Given cos (x + π / 12) = 1 / 3, what is the value of sin (2x - π / 3)?

Given that cos (x + π / 12) = 1 / 3, then:
cos(2x+ π/6)=2cos²(x+π/12)-1=2/9 - 1=-7/9
So:
sin(2x-π/3)=sin[(2x+ π/6)- π/2]
=-sin[π/2 -(2x+ π/6)]
=-cos(2x+ π/6)
=7/9

Given that the minimum value of function f (x) = - Sin ^ 2x + 2asin (x - π / 2) is a ^ 2 + 4a, find the value of real number a

f(x)=-sin^2x+2asin(x-π/2)
=-(1-cos^2x)+2a(-cosx)
=cos^2x-2acosx-1
=(cosx-a)^2-a^2-1
Due to - 1

Given the function f (x) = 2cos ^ 2x + 2 * root 3cosx + A, if the sum of the maximum and minimum value of F (x) on [- π / 6, π / 3] is 3, it is the value of real number a

If (cosx + 3 / 2) ^ 2 ≥ 0, then when (cosx + 3 / 2) ^ 2 = 0, when (cosx + 3 / 2) ^ 2 = 0, when (cosx x + 3 / 2) ^ 2 = 0, f (x) gets the minimum value, f (x) min = A-3 / 2 when x = 0, (cosx + 3 / 2 / 2) ^ 2 = 0, when x = 0, (cosx + 3 / 2 / 2) ^ 2 takes the maximum value, f (x) gets the maximum value, f (x) gets the maximum value, f (x) gets the maximum value, f (x) max = f (0) = a (x) max = f (0) = a (x) max = f (0) = a (x) max = f (0) = a (x) max = f (x) max = f (x) + 2 + 2 √ 3