Given the terminal point P (1, - 2) of angle α, find sin α, sin 2 α, cos 2 α

Given the terminal point P (1, - 2) of angle α, find sin α, sin 2 α, cos 2 α

sinα=y/r=-2/√(1^2+(-2)^2)=-2/√5=-2√5/5
cosα=x/y=1/√5=√5/5
sin2α=2sinαcosα=-4/5
cos2α=2cos^2α-1=2/5-1=-3/5
If you don't understand, please ask

Given that P (3,4) is a point on the terminal edge of angle 2, find sin2, Cos2, tan2

From point P (3,4)
We get x = 3, y = 4, r = 5
So sin2 = Y / r = 4 / 5,
cos2=x/r=3/5,
tan2=y/x=4/3

Given that there is a point P (sin2 / 3 π, Cos2 / 3 π) on the terminal edge of angle a, then the value of minimum positive angle a is

That is p (√ 3 / 2, - 1 / 2)
In the fourth quadrant
tana=y/x=-√3/3
So a = 2 π - π / 6 = 11 π / 6

In the first quadrant of the plane rectangular coordinate system, there is a point P (x, y), r = | op | = √ X & sup2; + Y & sup2;. If OP forms an acute angle α with the positive direction of X axis, then sin
α=( ),cosα=( ),tanα=( ).

sinα=( y/√（x^2+y^2)),
cosα=(x/√（x^2+y^2)),
tanα=(y/x).

It is known that the coordinates of ABC are a (3,0) B (0,3) C (COS α, sin α), α ∈ (π / 2,3 π / 2), if the vector AC is multiplied by the vector BC = - 1
Find (2Sin & sup2; α + 2Sin α * cos α) / (1 + Tan α)

AC = (COS α - 3, sin α), BC = (COS α, sin α - 3), ac * BC = cos α (COS α - 3) + sin α (sin α - 3) = 1-3 (COS α + sin α) = - 1, ｜ cos α + sin α = 2 / 3, square 1 + 2cos α sin α = 4 / 9, ｜ (2Sin & sup2; α + 2Sin α * cos α) / (1 + Tan α) = 2cos α sin α = - 5 / 9

It is known that the coordinates of ABC are a (3,0) B (0,3) C (COS α, sin α), α ∈ (π / 2,3 π / 2) respectively. If the vector AC is multiplied by the vector BC = - 1, then (2Sin & S) can be obtained
Find (2Sin & sup2; α + 2Sin α * cos α) / (1 + Tan α)

Vector AC = (cosa-3, Sina) vector BC = (COSA, sina-3)
So: cosa (cosa-3) + Sina (sina-3) = - 1
That is, Sina + cosa = 2 / 3, and the square of both sides is 2sinacosa = - 5 / 9
And (2Sin & sup2; α + 2Sin α * cos α) / (1 + Tan α) = 2sinacosa
So (2Sin & sup2; α + 2Sin α * cos α) / (1 + Tan α) = - 5 / 9

In △ ABC, vector AC = (1 + cos α, sin α), BC = (COS α, 1 + sin α), α ∈ (0, π / 2)
1. Find the size of │ ab │ and ∠ C;
2. Find the maximum area s of △ ABC

1. Vector AB = ac-bc = (1, - 1),
∴|AB|=√2.
2.|AC|=√(2+2cosα),
AB*AC=1+cosα-sinα,
cosA=AB*AC/(|AB||AC|)=(1+cosα-sinα)/[2√（1+cosα)],
(cosA)^2=(1+cosα-sinα-sinαcosα)/[2(1+cosα)]
=(1-sinα)/2,
sinA=√[(1+sinα)/2],
S=(1/2)|AB||AC|sinA
=√[(1+cosα)(1+sinα)/2],α∈(0,π/2),
Let t = sin α + cos α, then sin α cos α = (T ^ 2-1) / 2, t ∈ (1, √ 2],
S=√{[1+t+(t^2-1)/2]/2}
=(t+1)/2,
When t = √ 2, s takes the maximum value (√ 2 + 1) / 2

The linear C1 x = 1 + TCOS α y = Tsin α is known
The known straight line C1 x = 1 + TCOS α
Y = Tsin α (t is the parameter),
C2 x=cosθ
Y = sin θ (θ is the parameter), when α = π / 3, C1 / C2 intersection coordinates
I want to know the detailed steps. I know pcosx = x psinx = y

Taking α = π / 3 into C1, x = 1 + TCOS α, y = Tsin α, we can get x = 1 + T / 2, y = two-thirds root sign three T 1) from C1 / C2 intersection coordinates, we can get x = 1 + T / 2 = cos θ 2) two-thirds root sign three T = sin θ 3) 2) square + 3) square, we can get (1 + T / 2) square + (two-thirds root sign three T) square = cos θ square + sin θ square

It is known that the parameter equation of the straight line is x = - 1-tsin π 7 / 6, y = 2 + TCOS π / 7, and the inclination angle is calculated!

(x+1)/(y-2)=-(sinπ7/6)/(cosπ/7)=0.55
tana=0.55
a=28.8 du

It is known that the parameter equation of the straight line is x = - 1-tsin π 6 / 7, y = 2 + TCOS π / 7, and the inclination angle is calculated