Given - π / 2 < x < 0, SiNx + cosx = 1 / 5, find cos2x

Given - π / 2 < x < 0, SiNx + cosx = 1 / 5, find cos2x

sinx+cosx=1/5
Square on both sides:
(sinx+cosx)^2=1/25
1+2sinxcosx=1/25
2sinxcosx=-24/25
Using the formula of double angle:
sin2x=-24/25
∵-π/2

Given that x belongs to (0, π / 2), SiNx cosx = √ 5 / 5, find (cos2x-2sin2x-1) / (1-tanx)

sinx-cosx=√5/5
And SiNx + cosx, we get SiNx = (2 and 5) / 5 cosx = (root 5) / 5
So sin2x = 4 / 5
cos2x=-3/5
tanx=2
By substituting, we get: (cos2x-2sin2x-1) / (1-tanx) = 16 / 5

If TaNx = 1 / 2, then (SiNx + cosx) 2 / (cos2x) is equal to 2 of (SiNx + cosx) 2 is square

The original formula = [(SiNx) ^ 2 + 2sinxcosx + (cosx) ^ 2] / [(cosx) ^ 2 - (SiNx) ^ 2] is the same as dividing (cosx) ^ 2
=[(tanx)^2+2tanx+1]/[1-(tanx)^2]
=(1/4+1+1)/(1-1/4)
=3

F (x) = 2cos square α / 2sinx sin α cosx SiNx when α = π, the minimum value is obtained Who of you will give me the answer within two hours, and I will add the score. In fact, I will calculate it, but it is different from what the teacher said. I want to verify it, Is 2cos squared α / 2 times SiNx, a / 2 is a whole that is F (x) = 2cos squared (α / 2) * SiNx sin α cosx SiNx There is x = 90 degrees, the title is wrong, sorry

You can describe it more accurately. There are many inconsistencies. You can have a look
Derivation, f (x) '= 2cos square α / 2cosx + sin α SiNx cosx
=(2cos square α / 2-1) cosx + sin α SiNx
=cosαcosx+sinαsinx
=cos(α-x)
The extremum is obtained when f (x) '= 0
α-x=π/2+kπ
α=（k+1)π

Given the function y = SiNx + cosx + 2sinxcosx, find the maximum value of function y

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

Find the range of y = (3-cosx) / (2 + SiNx)

The original formula can be written as
y=[3-cosx]/[2-(-sinx)]
According to the slope formula
k=(y2-y1)/(x2-x1)
We know
The value of y can be seen as
The slope value of the line passing through point a (2,3) and point B (- SiNx, COS)
A (2,3) is a fixed point and B (- SiNx, COS) is a moving point
Let y '= cosx
x'=-sinx
Then (y ') ^ 2 + (x') ^ 2 = (cosx) ^ 2 + (- SiNx) ^ 2 = 1
That is, the locus of the point is a circle with radius 1 at the origin
Then when AB is tangent to the circle, the slope of the line has a maximum or minimum value
(the straight line AB passes through point a, so the analytic formula is y = kx-2k + 3)
Use the formula of distance from point to line. (when tangent, the distance from center of circle to AB is equal to radius)
1 = (3-2k) / √ (1 + K ^ 2)
The solution is k = (6 + 2 √ 3) / 3 or (6-2 √ 3) / 3
therefore
The value range of the original function is [(6-2 √ 3) / 3, (6 + 2 √ 3) / 3]

Y = SiNx / (2 + cosx) Don't use derivatives to explain white points

Y = SiNx / (2 + cosx) y (2 + cosx) = SiNx 2Y + ycosx = SiNx SiNx SiNx ycosx = 2Y from the auxiliary angle formula of trigonometric function, we can know that | SiNx ycosx ә≤√ (1 + y |) so | 2Y ә≤√ (1 + y?) 4Y | 1 + y | 3Y | ≤ 1 - √ 3 / 3 ≤ y ≤√ 3 / 3, the value range of the function is [-

What is the range of y = (3-sinx) / (2 + cosx)? The answer is [(6-2 roots 3) / 3, (6 + 2 roots 3) / 3]

Let k = (3-sina) / (2 + COSA), then K is the slope of the line passing through two points a (2,3) and B (- cosa, Sina). Therefore, if B is on the unit circle, then the straight line Y-3 = K (X-2) of a has a common point with the unit circle, so the distance between the center of the circle (0,0) and the line kx-y + 3-2k = 0 is less than the radius r = 1, so | 0 -

Value domain of y = SiNx / [cosx-2]

I'll teach you a simple method: SiNx = ycosx-2y, that is, sinx-ycosx = - 2y1 ^ 2 + y ^ 2 > = (2Y) ^ 2 to solve the inequality, y ^ 2 = C ^ 2. If it is changed to y = SiNx / (sinx-2), you will do nothing but express SiNx in the form of Y, and then solve two inequalities in [- 1,1] according to SiNx

F (x) = 1 / 2 (SiNx + cosx) - 1 / 2 | SiNx cosx | find the range of F (x)

f（x）=1／2（sinx+cosx）-1／2|sinx-cosx|
The absolute value sign is needed to simplify the function
When x ∈ [2K π + π / 4,2k π + 5 π / 4], SiNx ≥ cosx
f(x)=1/2(sinx+cosx)-1/2(sinx-cosx)=cosx
x∈[2kπ+π/4,2kπ+5π/4],k∈Z
/ / cosx ∈ [- 1, √ 2 / 2], i.e., f (x) ∈ [- 1, √ 2 / 2]
When x ∈ (2k π - 3 π / 4,2k π + π / 4] K ∈ Z, SiNx is a new type of nonlinear system