It is known that the vertex of angle α is at the origin, and the starting edge is the positive half axis of X axis. If the end edge of angle α passes through the point P (√ 3, y), and sin α = - √ 2 △ 4, y ≠ 0 (1), judge the quadrant where angle α is located (2), and calculate the values of cos α and Tan α

It is known that the vertex of angle α is at the origin, and the starting edge is the positive half axis of X axis. If the end edge of angle α passes through the point P (√ 3, y), and sin α = - √ 2 △ 4, y ≠ 0 (1), judge the quadrant where angle α is located (2), and calculate the values of cos α and Tan α

From the definition of trigonometric function: sin α = Y / R
r=√(x²+y²)=√(3+y²)
So, Y / √ (3 + Y & # 178;) = - √ 2 / 4 (Note: y
（1）
Because sin is negative, y < 0, α is in the third and fourth quadrant. And because the abscissa of P is greater than 0, P is in quadrant one or four. So α is in the fourth quadrant.
（2）
Because α is in the fourth quadrant, the cos value is positive, cos α = √ (1-sin & # 178; α) = (√ 14) / 4
Because α is in the fourth quadrant, the value of Tan is negative, Tan α = sin α / cos α = (√ 7) / 7
（1）
Because sin is negative, y < 0, α is in the third and fourth quadrant. And because the abscissa of P is greater than 0, P is in quadrant one or four. So α is in the fourth quadrant.
（2）
Because α is in the fourth quadrant, the cos value is positive, cos α = √ (1-sin & # 178; α) = (√ 14) / 4
Because α is in the fourth quadrant, the value of Tan is negative, Tan α = sin α / cos α = (√ 7) / 7
sinα=－√2/4
Given that the vertex of angle a is at the origin, the starting edge is at the positive half axis of X axis, and the final edge passes through the point P (1, - 2), then cos (π + a) is
The final edge passes through the point P (1, - 2),
Then x = 1, y = - 2
r=|OP|=√(x²+y²)=√5
Using the definition of trigonometric function
cosa=x/r=1/√5=√5/5
∴ cos(π+a)=-cosa=-√5/5
Given that the first term of arithmetic sequence is A1, the tolerance is D, BN = 3an + 4b, then whether the sequence is arithmetic sequence or not
bn+1-bn=3an+1+4b-(3an+4b)
=3an+1-3an
=3d
So it's an arithmetic sequence with a tolerance of 3D~
Yes. Bn-b (n-1) = 3an + 4d-3a (n-1) - 4D = 3an-3a (n-1) = 3 (an-a (n-1) = 3D, 3D is a constant, so it is an arithmetic sequence.
Yes, because the arithmetic sequence formula of term a is: an = a1 + (n-1) d,
Because BN = 3an + 4b, substituting an into BN, BN = 3A1 + (n-1) * 3D + 4B can be obtained
Because the general formula of arithmetic sequence is an = a1 + (n-1) d
As long as we convert BN into the general form of one side, we can show that it is an arithmetic sequence
In BN, we regard 3A1 + 4B as B1, and 3D as the tolerance of BN, so we get
BN = (3A1 +... Expansion
Yes, because the arithmetic sequence formula of term a is: an = a1 + (n-1) d,
Because BN = 3an + 4b, substituting an into BN, BN = 3A1 + (n-1) * 3D + 4B can be obtained
Because the general formula of arithmetic sequence is an = a1 + (n-1) d
As long as we convert BN into the general form of one side, we can show that it is an arithmetic sequence
In BN, we regard 3A1 + 4B as B1, and 3D as the tolerance of BN, so we get
Bn=(3A1+4b)+(n-1)*3d
That's it. Put it away
Is the square of X equal to 0 a quadratic equation of one variable
Please explain why
X & sup2; = 0 is a quadratic equation of one variable
When x = 0, x square = 0,
If x is not equal to O and both sides divide x, then x = 0
To sum up, X has and has only one solution, x = 0
It has to be
The equation with an integer of degree 2 is called the equation with the highest degree of unknowns.
So this equation is a quadratic equation of one variable!
yes. There's only one unknown, so it's one dollar.
The unknowns are quadratic, so they are quadratic.
Given nonempty set a = {X / K ^ 2-2}
B={x/x^2-3kx-4k^20},
={(x-4k)(x+k)0},
={-k=-2k+6
k^2+2k-8>=0
K=2
Because k > 0, k > = 2
If a is not an empty set,
-k
asdasd
Finding the domain of the function y = √ cos (SiNx)
Let cos (SiNx) ≥ 0, so 2K π - (π / 2) ≤ SiNx ≤ 2K π + (π / 2) (K ∈ z) Or - 5 π / 2 ≤ SiNx ≤ - 3 π / 2 or - π / 2 ≤ SiNx ≤ π / 2 or 3 π / 2 ≤ SiNx ≤ 5 π / 2 or Because π = 3.14, the above result is Or - 7.85 ≤ SiNx ≤ - 4.71 or -
It is known that {an} is an arithmetic sequence whose first term is a and tolerance is 1, BN = 1 + Anan. If BN ≥ B8 holds for any n ∈ n *, then the value range of a is______ .
The general formula of {an} is an = a + n-1, ∵ BN = 1 + Anan = 1 + 1An = 1 + 1A + n − 1. ∵ BN ≥ B8 ∵ 1 + 1An ≥ 1 + 1a8, that is, 1An ≥ 1a8. The sequence {an} is an increasing sequence, and the tolerance is 1, ∵ A8 = a + 8-1 < 0, A9 = a + 9-1 > 0. At this time, 1a8 < 0 (n ≥ 8) when 0 < n < 8, there is also an < A8, that is, 1An ≥ 1a8, and the solution is - 8 < a < - 7, so the answer is It is (- 8, - 7)
”What are the A, B and C of the quadratic equation of one variable?
a=1 ,b=0,c=0
”What are the A, B and C of the quadratic equation of one variable?
a=1;b=0;c=0;
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”What are the A, B and C of the quadratic equation of one variable?
a=1;b=0;c=0;
Hello, I'm glad to answer for you. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask. If you are satisfied, remember to adopt it
If you have any other questions, please take this question and send it to me. Please understand. Thank you.
I wish you progress in your study
If the equation log0.5 ^ x = A / 1-A about X has a solution on the interval (0,1), then the value range of the real number a?
Ask for detailed explanation
A / (1-A) > 0
a(a-1)
If the function f (x) = cos2x + SiNx cosx TX increases monotonically on [0, π 2], then the value range of real number T is______ .
∵ the function f (x) = cos2x + SiNx cosx TX is monotonically increasing on [0, π 2]; the derivative f ′ (x) ≥ 0 of the function f (x) is constant on the interval [0, π 2], and f ′ (x) = - 2sin2x + cosx + sinx-t is obtained, so - 2sin2x + cosx + sinx-t ≥ 0 is constant on the interval [0, π 2], that is, t ≤ - 2sin2x + cosx + SiNx always holds for X ∈ [0, π 2]. Note that the function g (x) = - 2sin2x + cosx + SiNx is easy to obtain g (x) in [0, π 2] Therefore, the answer is: (− ∞, 2 − 2]