If the terminal edge of angle α is known to pass through points (3, - 4), then the value of sin α + cos α is______ .

If the terminal edge of angle α is known to pass through points (3, - 4), then the value of sin α + cos α is______ .

∵ given that the terminal edge of angle α passes through points (3, - 4), then x = 3, y = - 4, r = 5, ∵ sin α = yr = - 45, cos α = XR = 35, sin α + cos α = - 15, so the answer is: - 15
If the terminal edge of angle a is known to pass through point P (3, - 2), the values of sina and cosa are obtained
√﹙3²+2²）=√13
sina=-2√13=-2√13／13
cosa=3／√13=3√13／13
√﹙3²+2²）=√13
sina=-2√13=-2√13／13
cosa=3／√13=3√13／13
If the terminal edge of angle a passes through point P (1, - 2), then the value of cosa + sina is?
op=√(1+5)=√5
∴cosa=1/√5
sina=-2/√5
cosa+sina=-1/√5=-√5/5
cosa=x/r=1/√5
sina=y/r=-2/√5
cosa+sina=1/√5-2/√5=-1/√5 =-√5/5
In the sequence {an}, an = 4n-52, a1 + A2 + +An = an2 + BN, n ∈ n *, where a and B are constants, then AB equals ()
A. 1B. -1C. 2D. -2
Method 1: when n = 1, A1 = 32, 32 = a + B, ① when n = 2, A2 = 112, 32 + 112 = 4A + 2B, ② from ① and ②, a = 2, B = - 12, ab = - 1. Method 2: A1 = 32, Sn = n (a1 + an) 2 = 2n2-12n, and Sn = an2 + BN, a = 2, B = - 12, ab = - 1
4X square - 8x-3 = 0
x²+2x=3/4
x²+2x+1=3/4+1
(x+1)²=7/4
x+1=±√7/2
x=(-2-√7)/2,x=(-2+√7)/2
x²-2x=3/4
x²-2x+1=3/4+1
(x-1)²=7/4
x-1=±√7/2
x=(2-√7)/2，x=(2+√7)/2
It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, and A2 = - 5, S5 = - 20
If the first item A1 of the sequence is not set, and the tolerance is D, the equations are directly listed. I don't know whether it will be considered as an error
No need to set
Because the first term of the sequence is A1, of course
The tolerance expressed by D in arithmetic sequence is the default
S5=5a3=-20
a3=-4
d=a3-a2=-4+5=1
a1=a2-d=-6
an=a1+(n-1)d=-6+n-1=n-7
Let the tolerance be d
S5=5a1+10d=5(a1+2d)=5a3=-20
a3=-4
d=a3-a2=(-4)-(-5)=1
a1=a2-d=-5-1=-6
an=a1+(n-1)d=-6+n-1=n-7
The general formula of sequence {an} is an = N-7
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Let the tolerance be d
S5=5a1+10d=5(a1+2d)=5a3=-20
a3=-4
d=a3-a2=(-4)-(-5)=1
a1=a2-d=-5-1=-6
an=a1+(n-1)d=-6+n-1=n-7
The general formula of sequence {an} is an = N-7
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A1 does not need to be set, but tolerance D needs to be set. After all, if it's scientific, it will be changed and deducted.
Let the tolerance be d
S5=a1+a2+a3+a4+a5+a5=(a1+a5)+(a2+a4)+a3=5a3=-20
a3=-4
d=a3-a2=-4-(-5)=1
an=a1+(n-1)d=a2+(n-2)d=-5+1×(n-2)=n-7
The general formula of sequence {an} is an = N-7
The function f (x) = - 5x + SiNx defined on (- 1.1). If f (1-A) + the square of F (1-A) > 0, then the solution set of real number is
It is easy to know that f (x) is an odd function. If we seek its derivative, we can see that it decreases monotonically. If we want the second inequality to hold, 1-A + 1-A square
LIM (N2 + 1 / N + 1-an-b) = 0, find a, B
N2 + 1 / N + 1: n + 1 divided by N + 1
Is Lim n - > 0 or N - > infinity?
If it is 0:1-0-b = 0, B = 1. A arbitrary
If it is infinite: n-an-b = 0, a = 1, B = 0
1, O
(the square of 8x - x) - (*) = the square of 4x + 7x-5, if the bracket part is stained, the bracket part is
Square of 8x - X - (square of 4x + 7x-5)
=The square of 8x-x-4x-7x + 5
=The square of 8x - the square of 4x - x-7x + 5
=4X squared - 8x + 5
It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, and A2 = - 5, S5 = - 20. (I) find the general term formula of the sequence {an}; (II) find the minimum value of n which makes the inequality Sn > an hold
(1) Let the tolerance of {an} be d (2 points) a1 + D = - 55a1 + 10d = - 20, and A1 = - 6D = 1 So an = - 6 + (n-1) · 1 = N-7 So we can get the following results: 1 (9) let n (n − 13) 2 > n − 7, that is, n2-15n + 14 > 0 The solution is n ＜ 1 or n ＞ 14 and N ∈ n *, so n ＞ 14, so the minimum value of n is 15 (13 points)