Given the function f (x) = SiNx + sin (x + row / 2), X belongs to R (1) to find the minimum integral period of F (x) and (2) to find the maximum and minimum value of F (x) (3) If f (x) = 3 / 4, find the value of sin2x

Given the function f (x) = SiNx + sin (x + row / 2), X belongs to R (1) to find the minimum integral period of F (x) and (2) to find the maximum and minimum value of F (x) (3) If f (x) = 3 / 4, find the value of sin2x

F (x) = SiNx + sinxcos π / 2 + sin π / 2cosx = SiNx + cosx = √ 2Sin (x + π / 2) t = 2 π / 1 = 2 π f (x), the maximum value of √ 2 and the minimum value of √ 2 are - √ 2F (x) = 3 / 4f (x) = SiNx + cosx = 3 / 4, the square of both sides is 1 + 2sinxcosx = 9 / 16, so sin2x = 2sinxcosx = 9 / 16-1 = - 7 / 16

Given the function f (x) = SiNx + sin (x + π / 2), find the minimum positive period of F (x) and the maximum value of F (x). If f (x) = 3 / 4, find sin2x?

f(x）=sinx+sin(x+π/2)=sinx+cosx
Therefore, the minimum positive period of F (x) is 2 π
f^2(x）=(sinx+cosx)^2=1+sin2x
sin2x=-7/16

Given that the maximum value of the function y = - Sin ^ 2x + sinx-a is 2, find the value of a (for detailed procedures,

y=-sin^2x+sinx-a
y=-sin^2x+sinx-1/4+1/4-a
y=1/4-a-(sinx-1/2)^2
Because - (sinx-1 / 2) ^ 2 ≠ 0
The maximum value of Y is 1 / 4-a
1 / 4-A = 2
a=-7/4

F (x) = 1 + sin (2x) + 4 (SiNx + cosx) f(x)=1+sin(2x)+4(sinx+cosx) Find the minimum The answer is 2-4 √ 2

f(x)=1+sin(2x)+4(sinx+cosx)=sin^2x+cos^2x+2sinxcosx+4(sinx+cosx)=(sinx+cosx)^2+4(sinx+cosx)+4-4=(sinx+cosx+2)^2-4=[√2(√2/2*sinx+√2/2*cosx)+2]^2-4=[√2(sinxcosπ/4+cosxsinπ/4)+2]^2-4=[√2sin(x+π/4...

It is known that the minimum value of the function f (x) = sin (2x) - A (SiNx + cosx) is g (a)

f(x)=2sinxcosx-asinx-acosx
Let SiNx + cosx = t = (√ 2) sin (x + π / 4)
Then - √ 2 ≤ t ≤√ 2
f(x)=t^2-1-at=t^2-at-1=(t-a/2)^2-1-a^2/4
1) When - √ 2 ≤ A / 2 ≤√ 2, that is - 2 √ 2 ≤ a ≤ 2 √ 2
When t = A / 2, the minimum value is - 1-A ^ 2 / 4
2) When a / 2 > √ 2, i.e. a > 2 √ 2, t = √ 2, the minimum value is 1 - √ 2A
3) When a / 2

The known function y = 7 / 4 + SiNx sin ^ 2x Value range of

Let t = SiNx, t ∈ [- 1,1]
y＝－t^2+t+7/4
＝-(t-1/2)^2+2
∈〔－1/4,2]

(√(1+tanx)-√(1+sinx))/(x√(1+sin^2x)-x) X tends to 0, find the limit, I was wrong. Sin ^ 2x is the square of (SiNx)

(√(1+tanx)-√(1+sinx))/(x√(1+sin＾2x)-x)
Simultaneous rationalization of numerator and denominator
=(tanx-sinx)(x√(1+sin^2x)+x)/[(x²*sin^2x)(√(1+tanx)+√(1+sinx))]
=(tanx-sinx)(√(1+sin^2x)+1)/[(xsin^2x)(√(1+tanx)+√(1+sinx))]
Lim √ (1 + sin ^ 2x) + 1 = 2, Lim √ (1 + TaNx) + √ (1 + SiNx) = 2
=(tanx-sinx)/(xsin^2x)
=tanx(1-cosx)/(xsin²x)
Equivalent infinitesimal, TaNx ~ x
=(1-cosx)/sin²x
Lophida's law
=sinx/(2sinxcosx)
=1/(2cosx)
=1/2
The above procedure omits the sign Lim ～

The range of y = (1 + SiNx) / (sin ^ 2x SiNx + 2) As the title

(1 + SiNx) + 4 / (1 + SiNx) range: range: 1 + 1 + SiNx) + 4 / (1 + SiNx) - 3} = 1 / {(1 + SiNx) + 4 / (1 + SiNx) - 3} calculate first (1 + SiNx) + 4 / (1 + SiNx) range: (1 + SiNx) + 4 / (1 + SiNx) range: | (1 + SiNx) + 4 / (1 + SiNx) range: | (1 + SiNx) + 4 / (1 + SiNx) | (1 + SiNx) + 4 / (1 + SiNx) | (1 + SiNx | 1 + SiNx | 1 + SiNx | 1 + SiNx | 1 + SiNx | 4 = 4 (if and only if and only(...)

If y = sin ^ 2x SiNx, then the value range of function y?

Y=sin^2x-sinx
= (sinx-1/2)^2-1/4
-1≤sinx≤1
-3/2≤sinx-1/2≤1/2
0≤(sinx-1/2)^2≤9/4
-1/4≤(sinx-1/2)^2-1/4≤2
Range [- 1 / 4,2]

Find the value range of y = / SiNx / + / cosx / + sin ^ 4 (2x)

y=/sinx/+/cosx/+sin⁴(2x)
∵|sin(x+π/2)}+|cos(x+π/2)|+sin⁴[2(x+π/2)]
=|cosx|+|sinx|+sin⁴x
The period of function T = π / 2
We only need to study the value range on a closed interval of X ∈ [0, π / 2]
When x ∈ [0, π / 2], y = SiNx + cosx + sin ⁴ (2x)
∵(sinx+cosx)²=1+2sinccosx=1+sin2x
∴sin2x=(sinx+cosx)²-1
Let t = SiNx + cosx = √ 2Sin (x + π / 4) ∈ [1, √ 2]
∴sin2x=t²-1,sin⁴(2x)=(t²-1)⁴
∴y=t+(t²-1)⁴
∵t∈[1,√2] ∴t²∈[1,2]
⁴ and T are increasing functions
(can be derivative)
⁴ y = t + (t? - 1) ⁴ is an increasing function
When t = 1, Ymin = 1, and T = √ 2, ymax = √ 2 + 1
The value range is [1. √ 2 + 1]