If cos (A-75 °) = - 1 / 3 and a is the fourth quadrant angle, then sin (105 ° + a) =?

If cos (A-75 °) = - 1 / 3 and a is the fourth quadrant angle, then sin (105 ° + a) =?

Cos (A-75 °) = - 1 / 3 and a is the fourth quadrant angle
A-75 ° is the third quadrant angle
∴ sin(a-75°)

It is known that cos (15 ° + α) = 1 3, α is the first quadrant angle, and the value of COS (75 ° - α) + sin (α + 105 °) is calculated

From cos (15 ° + α) = 1
3, α is the first quadrant angle,
Sin (15 ° + α) = 2
Two
3，
cos（75°-α）+sin（α+105°）
=sin[90°-（75°-α）]+sin[90°+（15°+α）]
=sin（15°+α）+cos（15°+α）
=2
Two
3+1
Three
=2
2+1
Three

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

From the title, cos α = - 1 / 3, α is the second quadrant angle,  sin? α = 1-cos? α = 1-1 / 9 = 8 / 9

If cos (π / 4 - α) = 12 / 13 and π / 4 - α is the first quadrant angle, then the value of [sin (π / 2 - 2 α)] / [sin (4 / π + α)] is

The formula of sin (π / 4 - α) = 5 / 13 is sin [2 (π / 4 - α)] / cos [π / 2 - (π / 4 + α)] = 2Sin (π / 4 - α) cos (π / 4 - α) = 1

Given that sin α = 1 / 3 and α is the second quadrant angle, the values of cos α and Tan α are obtained

Sin α = 1 / 3, and α is the second quadrant angle
Cos α = - root sign (1-sin ^ 2 α) = - root sign (1-1 / 9) = - 2 root sign 2 / 3
Tan α = sin α / cos α = (1 / 3) / (- 2 root 2 / 3) = - root 2 / 4

If α is the second quadrant angle, what quadrant is the point P (sin (COS α), cos (sin α))?

Because: α is the second quadrant angle, so - 1 < cos α < 0, take cos α as angle (radian) in the fourth quadrant, so sin (COS α) < 0; and because α is the second quadrant angle, so 0 < sin α < 1, take sin α as angle (radian) in the first quadrant, so cos (sin α) > 0

Sin a + cos a = (1 / 5) angle a is the angle of the fourth quadrant As the title

Because sin a + cos a = (1 / 5), cos a = 1 / 5-sina,
If Sina ^ 2 + cos ^ 2 = 1, then Sina = 4 / 5, - 3 / 5, then cosa = - 3 / 5, 4 / 5
A is the angle of the second or fourth quadrant

If sin α + cos α = a (0

∵sinα+cosα=a
∴(sinα+cosα)²=a²
sin²α+2sinαcosα+cos²α=a²
2sinαcosα+1=a²
2sinαcosα=a²-1
sin2α=a²-1

It is known that cos (a) + sin (a) = A and 0

From 0 only

It is known that a = (COS α, sin α). B = (COS β, sin β), 0 < β < α < π If | vector a-vector B | = √ 2, prove vector a ⊥ vector B Let vector C = (0,1), if vector a + vector b = vector C, calculate the value of α β

(1) | vector a-vector B | = √ (COS α - cos β) ^ 2 + (sin α - sin β) ^ 2 = √ 2 the square of both sides leads to 1 + 1-2 (COS α cos β + sin α sin β) = 2, so cos α cos β + sin α sin β = 0, so the order of magnitude of vector a and vector B is equal to 0, so vector a ⊥ vector b (2) if vector a + vector b = vector C = (0