The coordinates of known points a ` B ` C 'are a (3,0) B (0,3) C (COS & sin @), @ belonging to (π / 2,3 π / 2). 1) if AC = BC, find the value of angle @? ➹ is a vector, & is a

The coordinates of known points a ` B ` C 'are a (3,0) B (0,3) C (COS & sin @), @ belonging to (π / 2,3 π / 2). 1) if AC = BC, find the value of angle @? ➹ is a vector, & is a

∵ AC = BC
We can get that point C is on the perpendicular of AB, that is, y = X
sin@=cosα α=arccos(sin@)∈R

Given (sin α + 4) cos α + 1 = 2, find the value of (COS α + 3) (sin α + 1)

If cosa is the denominator
(sin α + 4) cos α = 1 Sina + 4 = cosa is obviously impossible
So I think you should put in parentheses
(sinα+4)\(cosα+1)＝2
Multiply past Sina + 4 = 2cosa + 2
sina+2=2cosa
Two sides square Sin & # 178; a + 4sina + 4 = 4cos & # 178; a
Then, because Sin & # 178; a + cos & # 178; a = 1
So Sin & # 178; a + 4sin a + 4 = 4-4sin & # 178; a
The solution is Sina = 0 or - 4 / 5
Cosa = 1 or 3 / 5
Then (COS α + 3) (sin α + 1) = 4 or 18 / 25

If sin α cos α = 1 / 4 and α is the third quadrant angle, then sin α + cos α

Sina + cosa = root (Sina ^ 2 + cosa ^ 2 + 2sinacosa) = root (1 + 2sinacosa)
Because sin α cos α = 1 / 4 and α is the third quadrant angle, 1 + 2sinacosa = 3 / 2
The value of sina + cosa is negative
So sin α + cos α = radical (3 / 2) = (radical 6) / 2

Sin (π + a) = 1 / 2, and a is the third quadrant angle, find cos (π - a)

Because sin (π + a) = 1 / 2, so sin (π + a) = - Sina = 1 / 2, that is Sina = - 1 / 2, because a is the third quadrant angle, so cosa

Given cos = - 1 / 3, α is the third quadrant angle, and sin (α + β) = 1, find cos (2 α + β)

α is the third quadrant angle
Then sin α = - √ (1-cos & sup2; α) = - 2 √ 2 / 3
sin(α+β)=1,∴cos(α+β)=0
cos(2α+β) =cosc[α+(α+β)]
=cosαcos(α+β)-sinαsin(α+β)
=0+2√2/3x1=2√2/3

If sin (π + a) = 4 / 5 and a is the angle of the third quadrant, what is the value of COS (2 π - a)?
RT.
It's better to have a problem-solving process. Thank you

That is - Sina = 4 / 5
The third quadrant is cosa

Evaluation: 3sin220 °− 1cos220 ° + 64sin220 °

3sin220°−1cos220°+64sin220°=(3cos20°−sin20°)(3cos20°+sin20°)sin220°cos220°+64sin220°=4(32cos20°−12sin20°)(32cos20°+12sin20°)14sin240°+64sin220°=4sin40°sin80°14sin240°+64sin220°=32cos40°+64•1−cos40°2=32

Simplifying sin (π / 3 + α) + sin (π / 3 - α)
Such as the title

We can use sum difference product, but we need to use induced formula first
For example, they all become sin
Original formula = sin (π / 6 - α) + sin (π / 6 + α)
=2sin0.5*[(π/6-α)+(π/6+α)] cos0.5[(π/6-α)-(π/6+α)]
=2sin π/6 cosα
=cosα

Sin [(2n + 1) π - π / 3] reduction

Sin [(2n + 1) π - π / 3] = sin (π - π / 3) = sin π / 3 = radical 3 / 2
If n is an integer

cosa+2sina=√5,tana=

Let sin a = X
Then, 2x + √ (1-x * x) = √ 5
Move the term, square both sides, and get 5x * x-4 √ 5x + 4 = 0
Solve the equation and get x = 2 / √ 5
So cos a = 1 / √ 5
tan a=sin a/cos a=2