Find the maximum and minimum values of the square of the function y = 2cos + 3sinx - 1 in the ratio interval (0,5 / 6 π)

Find the maximum and minimum values of the square of the function y = 2cos + 3sinx - 1 in the ratio interval (0,5 / 6 π)

y=2(cosx)^2+3sinx -1
=-2(sinx)^2+3sinx+1
=-2(sinx-3/4)^2+17/8
SiNx = 3 / 4, y = 17 / 8
The interval is an open interval, so there is no minimum
One

Find the maximum and minimum value of function y = 2-4 / 3sinx cos ^ 2 (x)

Sin? X + cos? X = 1y = 2-4 / 3sinx cos ^ 2 (x) = 2-4 / 3sinx + sin? X-1 = sin? X-4 / 3sinx + 1, let t = SiNx  t ∈ [- 1,1] y = t? - 4 / 3T + 1 t ∈ [- 1,1] / / the function y = 2-4 / 3sinx cos ^ 2 (x) is equivalent to finding y = t? - 4 / 3T + 1 t

The minimum value of the function y = cos ^ x + 3sinx + 2 is

y＝cos²x+3sinx+2
＝1－sin²x+3sinx+2
＝－sin²+3sinx+3
＝－(sinx-3/2)²+21/4
Because - 1 ≤ SiNx ≤ 1, so:
When SiNx = - 1, the function y has the minimum value - 1;
(when SiNx = 1, the function y has a maximum value of 5)

The minimum value of the function y = 4-3sinx-cos ^ 2 (x)

y=4-3sinx-cos^2(x) = sin^2x -3sinx +3
This is a quadratic function about SiNx. The symmetric axis of the quadratic function is SiNx = 3 / 2, on the right side of SiNx's own range, so y is monotonically decreasing with respect to SiNx
So the minimum of Y is obtained at SiNx = 1 and the maximum at SiNx = - 1
Range [1,7]

Find the period of y = cos (2x + Π△ 3) function

T=2π/2=π
The period is π

How to find the period of the function y = cos π 2x

y=cos2x/π
T=2π/2/π=π^2

The period of the function y = cos (2x + 3 π) · sin (2x - π) is

solution
y=cos(2x+3π)·sin(2x-π)
=-cos2x*（-sin2x）
=1/2sin4x
Period T = 2 π / 4 = π / 2

The minimum value of the function y = cosx + √ 3sinx on the interval [0, π / 2] is?

y=2cos(x+π/6)
In the range of X + π / 6: [π / 6,2 π / 3], the function decreases monotonically in this range
Therefore, when x = π / 2, the minimum value y = - 1

Let a = (√ 3sinx, cosx), B = (cosx, - cosx), f (x) = a · B-1 / 2 (x ∈ R) (1) find the minimum value and the minimum positive period of the function f (x); (2) let the opposite sides of the inner angles a, B, C of the triangle ABC be a, B, C respectively, and C = √ 3, f (c) = 0, if SINB = 2sina, find the values of a and B This is mainly the second sub topic,

If f (x) = 2A * B-1 = 2 (1 = (3) SiNx cosx + cos (x), 1 = (√ 3) sin2x + [1 + cos2x] - 1 = 2 [(3 / 3 / 2) sin2x + (1 / 2) cos2x] = 2 [sin2xcos (π / 6) + cos2xsin (π / 6)] = 2Sin (2x + π / 6), then: F (x) = 1 is 2Sin (2x + π / 6) = 1sin (2x + π / 6) = 1sin (2x + π / 6) = 1sin (2x + π / 6) = 1sin (2x + π / 6) = 1sin (2x + π / 6) = 1sin (2x + π / 6) = 1sin (2x + π / 6) = 1 because x ∈ [π

If - π / 2 ≤ x ≤ π / 2, then the maximum and minimum values of the function f (x) = cosx - √ 3sinx are?

It can be concluded from the meaning of the title
f(x)=cosx-√3sinx=2sin(x-π/3)
And - π / 2 ≤ x ≤ π / 2
So - 5 π / 6 ≤ X - π / 3 ≤ π / 6
So the maximum value of F (x) is 1 and the minimum value is - 2