Given that P (Tana, COSA) is in the third quadrant, then which quadrant is angle a in

Given that P (Tana, COSA) is in the third quadrant, then which quadrant is angle a in

The second quadrant is Tana
Given that the point P (Tana, COSA) is in the third quadrant, then which quadrant does the angle a change in?
Quadrant 2
Tana "0" shows that a is in the second and fourth quadrant
cosa
FX = SiNx + cosx when x ∈ [0, Π] find the maximum value of FX
Please show your skills to help me
Also, find the value of X at this time!
fx=sinx+cosx
=√2(√2/2*sinx+√2/2*cosx)
=√2sin(x+π/4)
∵x∈〔0,π〕
When x + π / 4 = π / 2, that is, x = π / 4, sin (x + π / 4) = 1 gets the maximum
So the maximum value of the original function is √ 2
FX = SiNx + cosx = ± √ 1 + sin2x x ∈ [0, Π], so the maximum value is √ 2
sin2x=1
x=π/4
fx=sinx+cosx=√2sin（x+π/4)
X ∈ [0, Π] where x = π / 4
So the maximum value is √ 2
You choose one of the two methods
A = {real number}, B = {irrational number}, find the difference set of a and B
Rational number
The maximum value of F (x) = sinx-2 / cosx + 1 is
The maximum value of F (x) = sinx-2 / cosx + 1 is f (x) = sinx-2 / cosx + 1 = sinx-2 / cosx - (- 1) let P (cosx, SiNx) Q (- 1,2) p be the slope k of PQ line on the unit circle, and the linear equation L is Y-2 = K (x + 1) so that l is tangent to the unit circle, then r = 1 = D = | K + 2 | / √ 1 + K ^ 2, k = - 3 / 4L: 3
The maximum value of SiNx is 1 and the minimum value is - 1,
-The maximum value of 2 / cosx is positive infinity,
When SiNx is closer to - 1, - 2 / cosx tends to be positive infinity,
So the maximum of F (x) is positive infinity,
In the title, sinx-2 and cosx + 1 are not bracketed. How can they be (sinx-2) / (cosx + 1)???
Is this still a default?
It is easy to know that the point P (cost, Sint) is any point on the unit circle X & sup2; + Y & sup2; = 1, where t ∈ R. the meaning of the function f (x) = (sinx-2) / (cosx + 1) is the slope of the line L passing through the fixed point Q (- 1,2) and any point P (cosx, SiNx) on the unit circle. Let L: Y-2 = K (x + 1), that is, l: kx-y + K + 2 = 0. When l is tangent to the unit circle, the distance from L to the center of the circle (0,0) is the semi expansion of the unit circle
It is easy to know that the point P (cost, Sint) is any point on the unit circle X & sup2; + Y & sup2; = 1, where t ∈ R. the meaning of the function f (x) = (sinx-2) / (cosx + 1) is the slope of the line L passing through the fixed point Q (- 1,2) and any point P (cosx, SiNx) on the unit circle. Let L: Y-2 = K (x + 1), that is, l: kx-y + K + 2 = 0. When l is tangent to the unit circle, the distance from L to the center of the circle (0,0) is the radius 1 of the unit circle. The function f (x) max = - 3 / 4
When x = 2n π + 3 / 4 π has a maximum and X ≠ 2n π + 1 / 2 π, the maximum is 3. (it's best to draw trigonometric function)
a. Why is a + (the square root of B) irrational when B is a real number
Many, such as B = 2, 3, 5, 7, 8 All right, a, whatever
a≠-b
A is any real number
B > 0 and as long as it is not square
When a is an irrational number, or B is not a square number and b > = 0, a + (the square root of B) is an irrational number
Finding the maximum value of the function y = a (SiNx) ^ 2 + cosx + 1 (0 ≤ x ≤ π)
I set cosx = t to solve it again. The classification of a is too boring. Is my method wrong? It's a complete process to find the great God
Can we do this? Y = a (SiNx) ^ 2 + cosx + 1 = a (1-cos-178; 178; x) + cosx + 1 and cosx + 1 let cosx = t = a-at \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\#178;) - (1 + 4A & #178;)
In the following real numbers, irrational numbers are: A. - 1 / 3 B. π C. √ 16 D. - 22 / 7
Irrational number refers to the number that can not be expressed in fractional form, that is, infinite acyclic number
Given sinxcosx = 3 / 8 and X ∈ (π / 4, π / 2), then cosx SiNx is equal to
x∈(π/4,π/2)
cosx
x∈(π/4,π/2)
cosx
If a and B are opposite to each other (AB is not equal to 0), C and D are reciprocal to each other, the absolute value of K is 2
What is the value of (2a + 2b) to the power of 2008 - (CD) to the power of 5 * (k to the power of 2 + 6k-1) - A / b
If a and B are opposite numbers (AB is not equal to 0), then a + B = 0, a / b = - 1
c. If D is reciprocal to each other, then CD = 1
If the absolute value of K is 2, then K ^ 2 = 4, k = 2 or - 2
(2a + 2b) to the power of 2008 - (CD) to the power of 5 * (k to the power of 2 + 6k-1) - A / b
=0 - 1*(4 + 6k -1) +1
= -2 - 6k
=- 14 or 10
What are you asking for? It's not a question