Given that Tan α and Tan β are the two real roots of the equation x2-5x + 6 = 0, we can find the value of 2sin2 (α + β) - 3sin (α + β) cos (α + β) + Cos2 (α + β)

Given that Tan α and Tan β are the two real roots of the equation x2-5x + 6 = 0, we can find the value of 2sin2 (α + β) - 3sin (α + β) cos (α + β) + Cos2 (α + β)

Solution 1: according to Weida's theorem, Tan α + Tan β = 5, Tan α· Tan β = 6, so tan (α + β) = Tan α + Tan β 1 − Tan α· Tan β = 51 − 6 = − 1. The original formula = 2sin2 (α + β) - 3sin (α + β) cos (α + β) + Cos2 (α + β) sin2 (α + β) + Cos2 (α + β) = 2tan2 (α + β) − 3T

In △ ABC, if Sina / 2 = cos (a + b) / 2, then △ ABC must be

A / 2 + (a + b) / 2 = 90 degrees
A + B / 2 = 90 degrees
Because a + B + C = 180 degrees
So c + B / 2 = 90 degrees
A=C
It is an isosceles triangle

In the triangle ABC, if Sina = (radical 3) / 2, then cos (2) - a =?

Triangle ABC
Sina = (radical 3) / 2
A = 60 ° or 120 ° is indicated
Cos (2-A) = cos (a) (induction formula)
When a = 60 degree
cosA=1/2
When a = 120 degree
cosA=-1/2

In the triangle ABC, if sin a / 2 = cos (a + b) / 2, then the triangle must be a triangle

SIN(A/2)=COS[(A+B)/2]=SIN[π/2-(A+B)/2]=SIN(C/2)
So a = C
Because sin (a) / a = sin (c) / C
So a = C
an isosceles triangle

Tan α / (Tan α - 1) = - 1 to find (sin α - 3cos α) / (sin α + cos α)

It's equal to - 1 / 3
From Tan α / (Tan α - 1) = - 1, Tan α = 1 / 2 is obtained
The denominator of (sin α - 3cos α) / (sin α + cos α) is divided by cos α,
　　=(tanα-3)/(tanα+1)
　　=-5/3

Two different signs of quadratic equation x + (M-3) x + M = 0, the absolute value of negative root is smaller than that of positive root, so the value range of real number m can be obtained

﹙-3m ／2﹢3﹚m／2＝m ﹣m／4﹢3m／2＝m -m﹢6m＝4m -m﹢2m＞0 m＞0 m＜2 ∴0＜m＜2

It is known that the two roots of the quadratic equation AX ^ 2 - (3a-1) x + a = 0 about X are both positive numbers. The value range of a is obtained

a≠0（1）
Discriminant = (3a-1) &# 178; - 4A & # 178; ≥ 0
9a²-6a+1-4a²≥0
5a²-6a+1≥0
(5a-1)(a-1)≥0
A ≥ 1 or a ≤ 1 / 5 (2)
Weida theorem
x1+x2=(3a-1)/a>0
a(3a-1)>0
a> 1 / 3 or A0
To sum up: so a ≥ 1 or a

Reduction of COS (- 2 α) / [2Sin & sup2; (π / 4 + α) * Tan (π / 4 - α)]
The answer is 1

sin(π/4+α)=cos[π/2-(π/4+α)]=cos(π/4-α)
tan(π/4-α)=sin(π/4-α)/cos(π/4-α)
So the denominator = 2cos & sup2; (π / 4 - α) * sin (π / 4 - α) / cos (π / 4 - α)
=2sin(π/4-α)cos(π/4-α)
=sin[2(π/4-α)]
=sin(π/2-2α)
=cos2α
Molecule = Cos2 α
So the original formula is 1

Tana satisfies the equation Tan & sup2; a-4tana + 4 = 0, find the value of 2Sin & sup2; a + sinacosa + cos & sup2; a

Tan & # 178; a-4tana + 4 = (tana-2) &# 178; = 0, Tana = 2, so Sina = 2cosa
2sin²a+sinacosa+cos²a=8cos²a+2cos²a+cos²a=11cos²a=11(cos2a+1)\2
Because of the universal formula cosx = [1-tan & # 178; (x / 2)] / [1 + Tan & # 178; (x / 2)], cos2a = - 3-5
It is found that 11 (cos2a + 1) 2 = 11-5

Given Tan & sup2; a = 2tan & sup2; B + 1, prove Sin & sup2; B + 1 = 2Sin & sup2; a

Proof: let x = (sin a) ^ 2,
y =(sin b)^2,
Then (COS a) ^ 2 = 1-x,
(tan a)^2 =x /(1-x).
Similarly,
(tan b)^2 =y /(1-y).
It is known that,
x /(1-x) =2y /(1-y) +1,
So x (1-y) = 2Y (1-x) + (1-x) (1-y),
That is, X - xy = 2Y - 2XY + 1 - Y - x + XY,
So y + 1 = 2x
That is, (sin b) ^ 2 = 2 (sin a) ^ 2 + 1
= = = = = = = = =
Exchange method
Pay attention to the number of times
X = (sin a) ^ 2, so (Tan a) ^ 2 = x / (1-x) is the key