If Sina = 2 / Radix 5, and cosa is less than 0, find the value of real number K

If Sina = 2 / Radix 5, and cosa is less than 0, find the value of real number K

∵ Sina = 2 / Radix 5, sin2a + cos2a = 1, cosa < 0
ν cosa = - 1 / root 5
∴k=tana=sina/cosa=-2.

If sin α = 2 / √ 5, and cos α

k=-2
The sin value is known, cos is negative, so it is obtuse angle
The line is y = - 2x
k=-2

We know that sin θ and cos θ are two real roots of the equation x-kx + K + 1 = 0 with respect to x, and 0

1=sin^2a+cos^2a
=(sina+cosa)^2-2sinacosa
=k^2-2(k+1)
=k^2-2k-2
k^2-2k-3=0
(k-3)(k+1)=0
K = 3 or - 1
When k = 3, x ^ 2-3x + 4 = 0, △ < 0, so omit
When k = - 1, x ^ 2 + x = 0
X = 0 or - 1
When Sina = 0 and cosa = - 1, a = π
When Sina = 3, π = 1
So k = - 1 a = π or 3 π / 2

In polar coordinates, given the circle ρ = 2cos θ and the straight line 3 ρ cos θ + 4 ρ sin θ + a = 0, find the value of real number a

If the circle ρ = 2cos θ and the straight line 3 ρ cos θ + 4 ρ sin θ + a = 0 have solutions at the same time, then
3ρcosθ+4ρsinθ+a
=3cosθ*2cosθ+4sinθ*2cosθ+a
=6*(cosθ)^2+8*sinθ*cosθ+a
=3[1+cos(2θ)]+4sin(2θ)+a
=3 + 5sin (2 θ + a) + a (let TGA = 4 / 3)
=0
Then a = - 3-5sin (2 θ + a) = - 3-5sin [2 θ + arctg (4 / 3)]

It is known that sin θ and cos θ are the two roots of the equation x 2 - √ MX (x is not in the root sign) + 1 / M = 0. Find the values of real numbers θ and m

Sin θ + cos θ = √ m squared
sin^2 θ +2sinθcosθ +cos^2 θ=m
1+2sinθcosθ=m
sinθcosθ=1/m
∴1+2/m=m
m^2-m-2=0
(m-2)(m+1)=0
M = 2 or M = - 1 (omitted)
When m = 2
sinθ+cosθ=√2
sinθcosθ=1/2
θ=2kπ+π/4

Sin α + sin β = root 2 / 2, find the value range of cos α + cos β

Sin α + sin β = square root of two sides sin ^ 2 α + sin ^ 2 β + 2Sin α sin β = 1 / 2 ① Let cos α + cos β = x, both sides square cos ^ 2 α + cos ^ 2 β + 2cos α cos β = x ^ 2 ② 1 + 1 + 2 sin α sin β + 2cos α cos β = 1 / 2 + x ^ 2, cos (α - β) = 1 / 2

Given sin α + sin β = (2 under the radical) / 2, find the value range of cos α + cos β? I don't know how to find the value range of the problem, the best solution process, thank you

Let cos α + cos β = m ① and sin α + sin β = √ 2 / 2 ②. ① 2 + ② 2: 2 + 2cos (α + β) = 1 / 2 + M2 cos (α - β) = 1 / 2m? - 3 / 4 ∵ - 1 ≤ cos (α - β) ≤ 1, ∵ - 1 ≤ 1 / 2m? - 3 / 4 ≤ 1, the solution is: - √ 14 / 2 ≤ m ≤√ 14 / 2, so - √ 14 / 2 ≤ cos α + cos β ≤√ 14 /

The value range of sin θ root sign (2) / 2

sinθ 2kπ-7π/6 2kπ-π/4

It is known that the angles of edges a, B and C of △ ABC are a, B, C respectively, and satisfy the requirements of cos? A + cos? B-cos? C = 1-sinasinb (1) Find the size of angle c (2) If the area of △ ABC is 4 √ 3, find the minimum value of a + 2B and the corresponding angle A

From 1-cos 2 a = sin? A, 1-cos? B = sin? B, 1-cos? C = sin? C, it is brought to the left; 1-sin? A-SiN? B + sin? C = 1-sinasinb; the two sides are simplified, and then by the sine theorem, ab = a? + B? - C? By the cosine theorem, cos C = a

In △ ABC, the opposite sides of angles a, B and C are a, B, C, a = 2, root 3, Tan ((a + b) / 2) + Tan (C / 2) = 4, SINB * sinc = cos ^ 2 (A / 2) Let a, B, and B, C. ^ 2 be the square

tan[（A+B）/2]+tanC/2=4,tan(A+B)=tan(π-C）
tan[（A+B）/2]+tanC/2=tan[π/2-C/2]+tanC/2=cot(C/2)+tan(C/2)
=Cos (C / 2) / sin (C / 2) + sin (C / 2) / cos (C / 2) = [sin (C / 2) ^ 2 + cos (C / 2) ^ 2] / (sinc / 2) (COSC / 2) = 2 / sinc = 4, C = π / 6 or 5 π / 6
cos(A/2)=cos[π-(B+C)]/2=sin[(B+C)/2]
cos(A/2)^2=sin[(B+C)/2]^2=[1-cos(B+C)}/2=sinBsinC,1-cosBcosC+sinBsinC=2sinBsinC
Cos (B-C) = 1, B-C = 0
B = C = π / 6,5 π / 6 should be rounded off (two obtuse angles are not allowed),