High school mathematics function calculation (to specific process) anxious ah tanα=3 4sinα-2cosα/5cosα+3sinα=? sinαcosα=? (sinα+conα)^2=?

High school mathematics function calculation (to specific process) anxious ah tanα=3 4sinα-2cosα/5cosα+3sinα=? sinαcosα=? (sinα+conα)^2=?

4sinα-2cosα/5cosα+3sinα
=(4tanα-2)/(5+3tanα)=10/14=5/7
sinαcosα
=sinαcosα/((sinα)^2+(cosα)^2)=tanα/((tanα)^2+1)=3/10
(sinα+conα)^2
=1+2sinαcosα
=8/5
4sin α - 2cos α / 5cos α + 3sin α = (4tan α - 2) / (5 + 3tan α) = 10 / 11 (divide by cos α)
Sin α cos α = sin α cos α / (sin ^ 2 * α + cos ^ 2 * α) = 1 / (Tan α + 1 / Tan α) = 1 / (3 + 1 / 3) = 3 / 10 (divide by sin α cos α)
(sinα+conα)^2=sin^2*α+cos^2α+2*sinα*cosα=1+2*3/10=8/5
Simultaneous equations: Sina square + cosa square = 1
sina/cosa=3，
The answer is: first, what's the matter with you? In parentheses, it's 5 / 7, the second 3 / 10,
The third 8 / 5,
Let FX = LG (2 / (1 + x) + a) be an odd function, then the value range of X with FX < 0 is
Odd function
f(-x)=-f(x)
f(-x)+f(x)=0
lg[2/(1+x)+a]+lg[2/(1-x)+a]=0
lg[2/(1+x)+a]*[2/(1-x)+a]=lg1
[2/(1+x)+a]*[2/(1-x)+a]=1
Multiply by (1 + x) (1-x)
(2+a+ax)(2+a-ax)=(1+x)(1-x)
(2+a)²-a²x²=1-x²
So (2 + a) ² = 1
a²=1
So a = - 1
f(x)=lg[2/(1-x)-1]
=lg(1+x)/(1-x)
Because f (x) is an odd function
So f (0) = LG (2 + a) = 0
so 2+a=1
so a=-1
so f(x)=lg(1+x)/(1-x)
Because FX > 0
so (1+x)/(1-x)>1
so 2x/(1-x)>0
x/(x-1)
Given the function f (x) = - 2asin (2x + π / 6) + 2A + B, X belongs to [π / 4,3 π / 4], whether there is a constant a and B belongs to Z, so that the range of F (x) is [- 3, radical 3-1]. If there is, find out the value of a and B; if not, explain the reason
x∈[π/4,3π/4]
2x+π/6∈[2π/3,5π/3]
So sin (2x + π / 6) ∈ [- 1, radical 3 / 2]
So - 2asin (2x + π / 6) ∈ [- a radical 3,2a]
So f (x) ∈ [2A radical 3A + B, 4A + b]
So 2A radical 3A + B = - 3, 4A + B = radical 3-1
The solution is a = 1, B = radical 3-5
Let f (x) LG (2 (1-x) + a) be an odd function, then the value range of X with F (x) < 0 is
A (- 1,0) B (0,1) C (negative infinity, 0) d (negative infinity, 0) U (1, positive infinity)
If f (x) is an odd function, a = - 1 can be obtained, and the range in brackets is 0 to 1
Given the function f (x) = 2Sin (ω x + φ), X ∈ R (where a > 0, ω > 0,0)
1、
T=2π/ω=π
ω=2
Over M
So - 2 = 2Sin (4 π / 3 + φ)
4π/3+φ=3π/2
φ=π/6
y=2sin(2x+π/6)
2、
Zero
ax^4+bx^3+cx^2+dx+e=3x^3+x-1=0*x^4+3x^3+0*x^2+1*x-1
If they are equal, the monomial coefficients of the same degree are equal
So a = 0, B = 3, C = 0, d = 1, e = - 1
Let f (x) = LG (2 / X-1 + a) be an odd function
Uuy
The function f (x) = 2Sin (ω x + φ) (ω > 0,0) is known
（1）ω=2,φ=π/3
(2) First move π / 6 to the left, then double the height
1) T = 2 * π / w = π; so w = 2;
So sin (π / 2 + φ) = 1 / 2, that is cos (φ) = 1 / 2; 0
F (x) = LG (- 1 + A / 2 + x) is an odd function, f (x)
F (x) = LG (- 1 + A / 2 + x) is an odd function, so when x = 0, f (x) = 0 - 1 + A / 2 = 1, a = 4
f(x)
There is a problem about the function of higher one. I hope you experts can give us some advice. We know that the quadratic function f (x) satisfies f (0) = 1 and f (x-1) - f (x) = 2x to find the analytic formula of F (x) ~ ~ ~ ~ please give us some advice. Thank you very much~~~~~~
Because f (x) is a quadratic function, so let f (x) = ax ^ 2 + BX + 1, and because that formula, generation, the corresponding terms are equal, you can solve the value of a, B, you can solve it, it is not very difficult, exercise the computing ability is very good, in the future, you must be optimistic about the conditions, such as quadratic function, this condition is very important
Meng
Let f (x) = ax ^ 2 + BX + C, because f (0) = 1, so C = 1
Because f (x-1) - f (x) = 2x, then a (x-1) ^ 2 + B (x-1) + 1-ax ^ 2-bx-1 = 2x expands to a = - 1, B = - 1, so f (x) = - x ^ 2-x + 1
The function f (x) defined on R has f (x + y) + F (X-Y) = 2F (x) f (y) for any X. y belonging to R, and f (x) is not equal to 0
Judging the parity of F (x)
(3) If there is a non-zero constant C such that f (C / 2) = 0, it is proved that f (x + C) = - f (x) holds for any x belonging to R. is f (x) a periodic function? Why?
By:
f(x+0)+f(x-0)=2f(x)f(0)
Namely:
2F (x) = 2F (x) f (0) and f (x) is not equal to 0
f(0)=1
f(0+x)+f(0-x)=2f(0)f(x)=2f(x)
f(-x)=f(x)
Even function