It is known that the line x × sin α + y × cos α + M = 0 is cut by the circle x ^ 2 + y ^ 2 = 2. The length of the line segment is four thirds and the root sign is three. Find the real number M rtrrrrrrrrrrrrrrrrrrrr

It is known that the line x × sin α + y × cos α + M = 0 is cut by the circle x ^ 2 + y ^ 2 = 2. The length of the line segment is four thirds and the root sign is three. Find the real number M rtrrrrrrrrrrrrrrrrrrrr

Distance from center of circle to straight line = | 0 + 0 + m | / √ (sin | + cos?) = | M|
String AB = 4 √ 3 / 3
Pass through the vertical line of seat O and foot C
Then OC = | M|
OA=r=√2
AC=1/2AB=2√3/3
OA^2=OC^2+AC^2
2=4/3+m^2
m=√6/3,m=-√6/3

Sin α + sin β = the root of 2, and find the value range of cos α + cos β Please say why A0 to 2 / 2 root 2 B-2 root 2 to 2 root 2 C-2 to 2 D-2 root 14 to 2 root 14

D
Sin α + sin β = root 2 of 2
Then (sin α + sin β) ^ 2 = 1 / 2 (1)
Let cos α + cos β = t
Then (COS) ^ 2 II.
① + 2
(sinα+sinβ)^2+(cosα+cosβ)^2=1/2+t^2
Expand to get
sinα^2+sinβ^2+2sinα*sinβ+cosα^2+cosβ^2+2cosα*cosβ=1/2+t^2
Organized
2+2cos(α-β)=1/2+t^2
t^2=3/2+2cos(α-β)
Because - 1 ≤ cos (α - β) ≤ 1
So 0 ≤ T ^ 2 ≤ 7 / 2
So the root of - 2 is 14 ≤ t ≤ 2 is 14
So the root of - 2 is 14 ≤ cos α + cos β ≤ 2

It is known that sin α + cos α = the root of 3 〈α﹤ π Find Cos2 α

Sin α + cos α = root 3 of 3
be
2sinacosa
=(sina+cosa)^2-(sin^2a+cos^2a)
=1/3-1
=-2/3
therefore
(cosa-sina)^2
=(cosa+sina)^2-4sinacosa
=1/3-2(-2/3)
=5/3
Because 0 〈α〈π
So π / 2 〈α﹤ π (2sinacosa)=_ 2/3

If sin α - sin β = 1-radical 3 / 2 cos α - cos β = - 1 / 2, cos (α - β) is equal to

(sinα-sinβ)=1-√3/2(sinα-sinβ)²=(1-√3/2)²sin²α-2sinαsinβ+sin²β=7/4 -√3 (1)cosα-cosβ=-1/2(cosα-cosβ)²=1/4cos²α-2cosαcosβ+cos²β=1/4 ...

In the triangle ABC, the opposite sides of the angle ABC are a, B, C, and B? 2 + C? 2 = a? 2 + root 3 BC, sinasinb = cos? C / 2 (1) Find the size of angle a, B, C (2) If the length of the center line am on the edge of BC is root 7, find the area of △ ABC The second is the main question

(1) According to the cosine theorem, we get a 2 = B 2 + C 2 - 2 bccosa. From the known B 2 + C 2 = a 2 + √ 3 BC cos a = √ 3 / 2A = π / 6 sinasinb = cos 2 (C / 2) sinasin B = (1 + COSC) / 22 sinasinb = 1-cos (a

It is known that a, B and C are the three inner angles of △ ABC, and their opposite sides are a, B and C respectively m＝(cosA 2，−sinA 2)， n＝(cosA 2，sinA 2) , and m• n＝1 Two (1) Find the value of angle A; (2) If a = 2 3, B + C = 4, find the area of △ ABC

(1) By
m•
n＝1
2, get cos2a
2−sin2A
2=1
2，
That is, cosa = 1
Two
∵ A is the internal angle of ᙽ ABC,
∴A=π
Three
(2) According to the cosine theorem: A2 = B2 + c2-2bccosa {A2 = (B + C) 2-3bc
That is 12 = 42-3 BC {BC = 4
3，
∴S△ABC=1
2bcsinA=1
2•4
3•
Three
2＝
Three
3．

In the triangle ABC, B = 60 degrees, cosa = 4 / 5, B = radical. 3. Find the value of sinc and the area of triangle ABC

Sine theorem a / Sina = B / Sinba = bsina / SINB = 6 / 5S △ ABC = (1 / 2) absinc = (1 / 2) * (6 / 5) * (√ 3) * (3 + 4 √ 3) / 102

In the triangle ABC, the opposite sides of the angle A.B.C are A.B.C, B = Pie / 3. Cosa = 4 / 5. B = radical 3. (1) find the value of sinc, (2) find the area of triangle ABC

∵ cosa = 4 ᙽ 5 leads to Sina = 3 / 5 SINB = (root 3) / 2 CoSb = 1 / 2
/ / sinc = sin (a + b) = sinacosb + sinbcosa = (4 * (root 3) + 3) / 10
∵ A / Sina = B / SINB, a = 6 / 5
ν s △ ABC = 1 / 2 * a * b * sinc = (72 + 54 * (root 3)) / 100

It is known that the opposite sides of the inner angles a, B and C of the triangle ABC are a, B and C respectively, and the root sign 3sinccosc cos square C = 1 / 2, If the vector M = (1, Sina) and vector n = (2, SINB) are collinear, find the value of A.B

[analysis] this paper mainly investigates the comprehensive application of the formula of double angle, auxiliary angle, sine formula of sum of two angles and trigonometric function of acute angle; (1) by using the formula of double angle and auxiliary angle, we can get sin (2c-30 °) = 1, combine the range of C to find C (2) from (1) C, we can get a + B, combined with vector collinear

In the triangle ABC, 1 / 2 + 2cosacosc = cos (A-C), (1) a + C = 4, the area of triangle ABC is (3 roots, 3 / 4), find B

1 / 2 + 2cosAcosC = cos(A-C)1 / 2 + 2cosAcosC = cosAcosC + sinAsinCcosAcosC - sinAsinC = - 1 / 2∴ cos(A+C) = - 1 / 2∵ A + C ∈ ( 0 ,π )∴ A + C = ( 2 / 3 )π∴ B = π - ( A + B ) = π / 3∴ sinB = ...