Solving a senior high school mathematics problem by trigonometric substitution Find the range of y = sqrt (x) + sqrt (3-3x)

Solving a senior high school mathematics problem by trigonometric substitution Find the range of y = sqrt (x) + sqrt (3-3x)

Let x = Cox ^ 2 (T)
The equation becomes y = cos (T) + sqrt (3) sin (T);
T belongs to [0, π / 2];
Then the sum difference formula of integration is used;
Y = 2cos (T - π / 3); the range of T - π / 3 is [- π / 3, π / 6]
The range of Y is [1,2]

A high school mathematics problem (triangle)
In the triangle ABC, the opposite sides of the internal angles a, B and C are a, B and C respectively. The known vector ca * vector CB = C ^ 2 - (a-b) ^ 2
(1) Find the value of COSC (2) if a is an obtuse angle, find the value range of SINB
Hope that the master gives a detailed answer, thank you

1. COSC = (vector ca * vector CB) divided by (/ Ca / * / CB /) = [C ^ 2 - (a-b) ^ 2] / (a * b)

High school mathematics (triangle)
In △ ABC, the edges of angles a, B and C are a, B and C respectively, and f (a) = 2cosa / 2Sin (π - A / 2) + Sin & # 178; a / 2-cos & # 178; a / 2
(1) Finding the maximum of F (a)
(2) If f (a) = 0, C = 5 π / 12, a = √ 6, find the value of B

f(A)=2cosA/2sin(π-A/2)+sin²A/2-cos²A/2
=2cosA/2*sinA/2-(cos²A/2-sin²A/2)
=sinA-cosA
=√2sin(A-π/4)
(1) The maximum value of F (a) = 2
(2)
f（A）=0
sin(A-π/4)=0
A is the inner corner
A=π/4
C=5π/12
B=π-π/4-5π/12=π/3
Application of sine theorem
a/sinA=b/sinB
√6/(√2/2)=b/(√3/2)
b=3

High school mathematics triangle
The range (α∈ R) of (√ 2 + sin α) / (√ 2 + cos α)? Process and result

The geometric meaning of (√ 2 + sin α) / (√ 2 + cos α) = [sin α - (- √ 2)] / [cos α - (- √ 2)] is: the slope of the line between the midpoint (COS α, sin α) and the point (- √ 2, √ 2) in the plane rectangular coordinate system, and because α∈ R is any real number, the composition set of the moving points (COS α, sin α) is the unit circle X & su

Triangle ABC. The opposite sides of angle A.B.C are a.b.c. s is the area of triangle, satisfying that s is less than 1 / 4 (B2 + c2-a2). ① the range of angle A. ② the range of F (a) = 1 + sinacosa cos ^ 2A

(1) Cosine theorem B & sup2; + C & sup2; - A & sup2; = 2bccosa
S = 0.5 * bcsina, so 0.5bcsina < 0.25 * 2bccosa, i.e. Tana < 1, i.e. 0 degree

You're bound to get it, triangle
1. Given that a, B and C are the three inner angles of a triangle, and lgsina-lgsinb-lgcosc = LG2, try to determine the shape of the triangle
2. Prove the square of TaNx + one of the squares of TaNx = [2 (3 + cos4x)] / 1-cos4x
3. Two of the equations 2x ^ 2 - (radical 3 + 1) x + M = 0 of X are sin θ and cos θ, θ∈ (0,2 π)
(1) The value of sin θ / (1-1 / Tan θ) + cos θ / (1-tan θ)
(2) The value of M
(3) Equation and the value of θ
If you can answer as many as you can, try your best to answer what others haven't answered,

sinA/sinB=2cosC
sinA=2sinBcosC
sin(B+C)= 2sinBcosC
sinBcosC+sinCcosB=2sinBcosC
sinCcosB-sinBcosC=0
sin(C-B)=0
C-B = 0 or π (rounding)
B＝C
sinA=2sinBcosC
sinA=sin2B
A = 2B or a + 2B = π
So △ ABC is an isosceles right angle or an isosceles triangle sin θ cos θ θ √∫△ a
To sum up, △ ABC is an isosceles triangle
3) The equation is solved by Weida theorem X1 + x2 = - B / a = - (- √ 3-1) / 2 = (√ 3 + 1) / 2
When θ = 30 °, sin θ = 1 / 2, cos θ = √ 3 / 2, Tan θ = √ 3 / 3
When θ = 60 °, sin θ = √ 3 / 2, cos θ = 1 / 2, Tan θ = √ 3
The first question is to directly substitute the value into the two cases
2) X1 * x2 = C / a = m / 2 = 1 / 2 times √ 3 / 2
That is, M / 2 = √ 3 / 4
∴ m=√3/2

Remember there is a kind of problem is: isosceles or right angle triangle line, someone still remember the original problem is how?

For isosceles, a = B (a-b = 0 °);
At right angles, a + B = 90 degrees;
If this is the case, it should be trigonometric function to determine the type of triangle
SIN（A-B）=0 OR COS（A+B）=0
That is: sin (a-b) * cos (a + b) = 0

In ABC, the edges opposite the three interior corners ABC are ABC
If B = radical 6 / 3C,
And B = 2C
Then the value of COSC is?

b/sinB=c/sinC
sinB
sinB=sin2C=2sinCcosC
Give you a hint!

1. In △ ABC, if a = 2, then bcosc + CoSb is equal to?
2. In △ ABC, if B = asinc, C = acosb, what kind of triangle is the triangle?
3. In △ we know that a-2b + C = 0, 3A + b-2c = 0 sinA:sinB Sinc is equal to?
4. In △ ABC, it is known that sin & sup2; a + Sin & sup2; b-sin asinb = Sin & sup2; C, and ab = 4
5. In △ ABC, the opposite sides of angle A.B.C are a, B, C respectively. If [(radical 3) b] cosa = acosc, then cosa is equal to?
6. In △ ABC, it is known that sina: SINB = radical 2:1, C & sup2; = B & sup2; + (radical 2) BC?
7. In △ ABC, the opposite sides of angle A.B.C are a, B, C. and a ＞ B ＞ C. If a & sup2; < B & sup2; < C & sup2;, then the value range of ∠ A is?
8. In △ ABC, the edges opposite the angle a.b.c. are a, B, C. then a: (B + C) + B: (a + C) is equal to?

In the third question, the relationship between a, B and C can be obtained by combining the two equations. Their ratio is the ratio of their sinusoidal value according to the sinusoidal theorem

On solving triangles
In △ ABC, the bisector CD of ∠ A: B = 1:2, and the triangle area is divided into two parts of 3:2, then cosa is equal to

Let a = x and B = 2x,
S△ ACD:S △BCD=0.5CA•CD•sin0.5C:0.5CB•CD•sin0.5C=3:2
∴AC:BC=3:2
AC/sinB=BC/sinA
∴AC/BC=sinB/sinA=sin2A/sinA=3/2
2sinA•cosA/sinA=3/2
COS∠A=3/4