The vertex of the known angle α coincides with the origin of the rectangular coordinate system, the starting edge is on the positive half axis of the x-axis, and the final edge passes through the point P (- 1,2). The values of sin α and COS (π + α) are calculated

The vertex of the known angle α coincides with the origin of the rectangular coordinate system, the starting edge is on the positive half axis of the x-axis, and the final edge passes through the point P (- 1,2). The values of sin α and COS (π + α) are calculated

We can get x = - 1, y = 2, r = 1 + 4 = 5, sin α = yr = 25 = 255. Cos (π + α) = - cos α = - XR = - − 15 = 55
The vertex of the known angle a coincides with the origin of the rectangular coordinate system. The starting edge is the positive half axis of the x-axis, and the final edge passes through the point P (- 3,4), so sin (2a + 2 π / 3) can be obtained
sina=4/5,cosa=-3/5
sin（2α+2π/3）
=sin2acos2π/3+cos2asin2π/3
=-1/2sin2a+√3/2cos2a
=-sinacosa+√3/2(cos^2a-sin^2a)
=12/25－√3/2*7/25
＝(24-7√3）/50
Given that the vertex of angle a coincides with the origin of rectangular coordinate system, the starting edge is on the positive half axis of X axis, and the ending edge is on y = - 2x, and X is less than or equal to 0, then sin (2a + 2 π|3) can be obtained
Give me an idea
Let {2} cosin a = {2} a = {2} cosin a = {2} a = {2} a = {2} A / sin a = {2} a = {2} a = {2} A / sin a = {2} a + 2} a on the edge of the second semiaxis
The square of x minus x minus 3 / 4 is equal to 0. How to use the matching method
I can't understand it
x^2-x-3/4=0
X ^ 2-2 * (1 / 2) * x-3 / 4 = 0 first becomes a linear term, and the term is in the form of 2 × (some number)
X ^ 2-2 * (1 / 2) * x + 1 / 4-1 / 4-3 / 4 = 0, plus the square of a number, minus
X ^ 2-2 * (1 / 2) * x + (1 / 2) ^ 2 = 3 / 4 + 1 / 4, the left side forms the complete square formula, and the right side is the constant
(x-1 / 2) ^ 2 = 1, the left is the complete square, and the right is 1
X1-1 / 2 = 1 get X1 = 3 / 2, the square root of the right number has positive and negative values
X2-1 / 2 = - 1 to get x2 = - 1 / 2
Help me! Let set a = {x | 1
B={x|x
The decreasing function f (x) defined in (- ∞, 3) such that f (A & sup2; - SiNx) ≤ f (a + 1 + cos & sup2; x) is a real number for all x, and the norm of a is obtained
The decreasing function f (x) defined in (- ∞, 3) such that f (A & sup2; - SiNx) ≤ f (a + 1 + cos & sup2; x) is a real number for all x, and the value range of a is obtained
A & sup2; - SiNx > = 3A + 1 + cos & sup2; x > = 3A & sup2; - SiNx > = a + 1 + cos & sup2; X simultaneous inequality system 3-A & sup2; > = SiNx if this is true for all x, then it must be greater than its maximum 3-A & sup2; > = 1. Similarly, using cos2x = 1-sin ^ 2x to simplify the two inequalities sin ^ 2 > = A-1 and - A ^ 2 + a
For this kind of topic, first, we must learn to establish the awareness of domain priority
a²-sinx≤3
a+1+cos²x≤3
a²-sinx》=a+1+cos²x
Join the above groups to solve the problem
For this kind of topic, first, we must learn to establish the awareness of domain priority
a²-sinx≤3
a+1+cos²x≤3
a²-sinx》=a+1+cos²x
Join the above groups to solve the problem
Interviewer: wengpaul - Jieyuan level 5 2009-9-20 17:38
Report a & sup2; - SiNx > = 3
A + 1 + cos & sup2; x >... Expansion
For this kind of topic, first, we must learn to establish the awareness of domain priority
a²-sinx≤3
a+1+cos²x≤3
a²-sinx》=a+1+cos²x
Join the above groups to solve the problem
Interviewer: wengpaul - Jieyuan level 5 2009-9-20 17:38
Report a & sup2; - SiNx > = 3
a+1+cos²x>=3
a²-sinx>=a+1+cos²x
System of simultaneous inequalities
3-A & sup2; > = SiNx if this holds for all x, then it must be greater than its maximum
3-a²>=1
In the same way
Using cos2x = 1-sin ^ 2x to simplify the last two inequalities
Sin ^ 2 > = A-1 and - A ^ 2 + A
Given that the sequence {an} satisfies an + 1 = (3an + 1) / (an + 3), A1 = - 1 / 3, it is proved that 1 / (an) + 1 is an arithmetic sequence and an is obtained
Wrong title retransmission
Given that the sequence {an} satisfies an + 1 = (an-1) / (an + 3), A1 = - 1 / 3, prove that 1 / (an) + 1 is the arithmetic sequence, and find an
a(n+1)=[a(n)-1]/[a(n)+3],
a(n+1)+1=[a(n)-1]/[a(n)+3] +1=[2a(n)+2]/[a(n)+3]=2[a(n)+1]/[a(n)+3],
If a (n + 1) + 1 = 0, then a (n) + 1 = 0,..., a (1) + 1 = 0, which is contradictory to a (1) = - 1 / 3
Therefore, a (n) + 1 is not 0
1/[a(n+1)+1] = [a(n)+3]/[2a(n)+2] = [a(n)+1+2]/[2a(n)+2] = 2/[2a(n)+2] + [a(n)+1]/[2a(n)+2]
=1/[a(n)+1] + 1/2,
{1 / [a (n) + 1]} is an arithmetic sequence with the first term of 1 / [a (1) + 1] = 1 / [1-1 / 3] = 3 / 2 and tolerance of 1 / 2
1/[a(n)+1] = 3/2 +(n-1)/2 = (n+2)/2,
a(n)+1=2/(n+2),
a(n)=2/(n+2) - 1 = -n/(n+2)
On the two roots of the quadratic equation (2a-1) x2 + (a + 1) x + L = 0 of X are equal, then a is equal to ()
A. 1 or 5B. - 1 or 5C. 1 or - 5D. - 1 or - 5
∵ the quadratic equation of one variable with respect to X has two equal roots of real numbers, and the solution is a = 1 or 5
Set a = {X / X5} B = {a
A greater than or equal to 5 or a less than or equal to - 5
adad
Sincox + function (F + s) / known
Let a α be the second quadrant angle and Tan (π - α) = 1 / 2, then find the value of F (α)
The 1 + is on top
1, domain of definition: X ≠ K π, K ∈ Z
2, because Tan (π - α) = 1 / 2, Tan α = - 1 / 2
f（x）=(1+cos(x+π/6) )/ sinx
=(√(sinx^2=cosx^2)+cosxcoxπ/6-sinxsinπ/6)/sinx
=√(1+cotx^2)+cotx*√3/2-1/2
Therefore, f (α) = √ 5 - √ 3-1 / 2
There are two main points in this problem, one is the conversion of 1, 1 = SiNx ^ 2 + cosx ^ 2, and the other is to pay special attention to the quadrant of the angle
The domain x ≠ K π, K is an integer
Function f (x) = [1 + cos (x + π / 6)] / SiNx
=[1+cosxcosπ/6-sinxsinπ/6]/sinx
=1/sinx+√3/2cotx-1/2
α is the second quadrant angle, and Tan α = - 1 / 2
cotα=-2
1+cot^2α=5=1/sin^2α
√5=1/sinx
Function f (α) = √ 5 - √ 3-1 / 2
KPI does not belong to the domain (z)
f(a)=1+cos(a+pi/6)/sina=1+(cosacospi/6-sinasinpi/60/sina （1）
tan(pi-a)=-tana=1/2
So Tana = - 1 / 2
A is the second quadrant angle, so sina is greater than 0 and cosa is less than 0
Because this problem can not determine the specific value of angle a, when we know that Tana = - 1 / 2, we might as well consider sin
Domain x is not equal to KPI (k belongs to Z)
f(a)=1+cos(a+pi/6)/sina=1+(cosacospi/6-sinasinpi/60/sina （1）
tan(pi-a)=-tana=1/2
So Tana = - 1 / 2
A is the second quadrant angle, so sina is greater than 0 and cosa is less than 0
Because this problem can not determine the specific value of angle a, when we know Tana = - 1 / 2, we might as well consider Sina = 1, cosa = - 2
Just substitute (1) directly
(because Tana = Sina / cosa is a proportional relationship, it has no effect on the result.)