Given cosa = 3 / 5 and the first quadrant angle of a, find the value of sina and Tana

Given cosa = 3 / 5 and the first quadrant angle of a, find the value of sina and Tana

sin a =4/5
tan a= 4/3
sinα=4/5,tanα=4/3
If the terminal edge of angle a is on the line y = - x, then what is Sina / [cosa] + [Sina] / cosa?
[represents the absolute value, not only the answer,
1. If a is the second quadrant angle, then a = 135 + 360K, K is an integer, so Sina > 0, cosa < 0, so the original formula = - 1 + (- 1) = - 2.2, where a is the fourth quadrant angle, then a = - 45 + 360K, K is an integer, so Sina < 0, cosa < 0, so the original formula = - 1 + (- 1) = - 2
Hope to adopt
Sina = 4 / 5, a is the second quadrant angle, find cosa, Tana, COTA
Because a is the second quadrant
therefore
cosa= -3/5
tana= -4/3
cota= -3/4
A ∈ (0, π / 2), then the minimum of (3 + 2sinacosa) / (Sina + COSA)
(3 + 2sinacosa) / (Sina + COSA) = (2 + 1 + 2sinacosa) / (Sina + COSA) = (2 + Sin & # 178; a + cos & # 178; a + 2sinacosa) / (Sina + COSA) = 2 / (Sina + COSA) + (Sina + cosa) ≥ 2 √ 2 / (Sina + COSA) * (Sina + COSA) = 2 √ 2
Let: Sina + cosa = t, then: t ∈ (1, √ 2]
(sina＋cosa)²=t²
1＋2sinacosa=t²
If 2sinacosa = T & # 178; - 1, then:
M=(3＋2sinacosa)/(sina＋cosa)
=(3＋t²－1)/(t)
=(2/t)＋t
The expansion function (t ∈ 2,} and ∈ t ∈ 2
Let: Sina + cosa = t, then: t ∈ (1, √ 2]
(sina＋cosa)²=t²
1＋2sinacosa=t²
If 2sinacosa = T & # 178; - 1, then:
M=(3＋2sinacosa)/(sina＋cosa)
=(3＋t²－1)/(t)
=(2/t)＋t
Because t ∈ (0, √ 2], and the function (2 / T) + T decreases in (0, √ 2], then:
The minimum value of M is obtained when t = √ 2, and the minimum value is 2 √ 2
(1 / 2) the function f (x) = - 5x + SiNx defined on (- 1,1). If f (1-A) + the square of F (1-A) > 0, then the value range of real number a is（
Because f (- x) = - f (x), the function f (x) is odd and monotone decreasing on (- 1,1)
f(1-a²)>-f(1-a)=f(a-1)
So - 1
F (x) = - 5x + SiNx is an odd function. f'(x)=-5+cosx0，f(1-a)>-f(1-a^2)=f(a^2-1)，1-a0，a1
-1
LIM (n →∞) (n ^ 2 / (an + b) - n ^ 3 / (2n ^ 2-1)) = 1 / 4 for a, B
The results are as follows
[(2-a)n^4-bn^3-n^2]/[(an+b)(2n^2-1)]
The highest numerator is four times and the highest denominator is three times,
If n is going to infinity, the coefficient of n ^ 4 in the molecule must be 0,
That is, a = 2
In this case, it is reduced to [- BN ^ 3-N ^ 2] / [(an + b) (2n ^ 2-1)]
Divide up and down by n ^ 3, [- B-2 / N] / [(a + B / N) (2-1 / N ^ 2)]
When n tends to infinity, the limit of 1 / N 1 / N ^ 2 is 0,
The above formula = - B / 2A = 1 / 4
b=-1
First, solve the following equation in the evaluation [- x quadratic + 5 + 4x] + [5x-4 + 2x quadratic] where x = - 2
The original formula = - X & # 178; + 5 + 4x + 5x-4 + 2x & # 178;
=x²+9x+1
=4-18+1
=-13
Original formula = - x quadratic + 5 + 4x + 5x-4 + 2x quadratic
=(2x quadratic - x quadratic) + (4x + 5x) + 5-4
=(2-1) x quadratic + (4 + 5) x + 5-4
=X quadratic + 9x + 1
If x = - 2, then the original algebra = (- 2) quadratic + 9 * (- 2) + 1
... unfold
Original formula = - x quadratic + 5 + 4x + 5x-4 + 2x quadratic
=(2x quadratic - x quadratic) + (4x + 5x) + 5-4
=(2-1) x quadratic + (4 + 5) x + 5-4
=X quadratic + 9x + 1
If x = - 2, then the original algebra = (- 2) quadratic + 9 * (- 2) + 1
=4-18+1
=-13. Put it away
=(-x+5)(x+1)+(x+1)(2x+3)-7
=(x+1)(x+8)-7
=-13
Let Sn be the sum of the first n terms of the arithmetic sequence {an}, S5 = 3 (A2 + A8), then the value of a5a3 is___ .
∵ {an} is an arithmetic sequence, ∵ S5 = a1 + A2 + +A5 = 5A3, A2 + A8 = 2a5, and S5 = 3 (A2 + A8), 〈 5A3 = 3 × 2a5, 〈 a5a3 = 56, so the answer is 56
If f (1-A) + F (1-A Square) > 0, then the value range of real number a is ()
It's like (1, √ 2). Let me tell you the way of thinking. We can find that the derivative is monotonic decreasing, and it's an odd function, and then we make an equation. I'm not good at computing
An=(-2-1/n2,1/n) lim n→∞ An=
N or the square of n →∞
No matter → + ∞ or → - ∞, the reciprocal of N or the square of n will approach 0
So the answer is an = (- 2,0)
An=(-2,0)