sin(alpha-bita)=1/2 and cos(alpha+bita)=1/2, where alpha and bita are positive acute angles then what is the value of alpha and bita
45&15
To find the value of α and β, we can use the trigonometric identities for the sum and difference of angles.
Given:
sin(α - β) = 1/2 --(1)
cos(α + β) = 1/2 --(2)
To simplify the problem, we'll rewrite equation (1) as sin(α)cos(β) - cos(α)sin(β) = 1/2.
Now, let's use the double angle identities for sine and cosine:
sin(α - β) = sin α cos β - cos α sin β
cos(α + β) = cos α cos β - sin α sin β
Plugging the given values into these identities, we have:
sin α cos β - cos α sin β = 1/2 --(3)
cos α cos β - sin α sin β = 1/2 --(4)
Next, let's use the Pythagorean identity: sin²x + cos²x = 1
Rearrange equation (3) to:
sin α cos β = 1/2 + cos α sin β
Now, square both sides of the equations (3) and (4):
(sin α cos β)² = (1/2 + cos α sin β)²
(cos α cos β)² = (1/2 - sin α sin β)²
Expanding these equations, we get:
sin² α cos² β = 1/4 + cos α sin β + cos² α sin² β + cos α sin β
cos² α cos² β = 1/4 + sin α sin β - cos² α sin² β - sin α sin β
Using the Pythagorean identity, sin²x = 1 - cos²x and cos²x = 1 - sin²x, we can rewrite the equations as:
(1 - cos² α) cos² β = 1/4 + cos α sin β + cos² α (1 - cos² β) + cos α sin β
(1 - sin² α) cos² β = 1/4 + sin α sin β - cos² α (1 - cos² β) - sin α sin β
Expanding these equations further:
cos² β - cos² α cos² β = 1/4 + cos α sin β + cos² α - cos² α cos² β + cos α sin β
cos² β - sin² α cos² β = 1/4 + sin α sin β - cos² α + cos² α cos² β - sin α sin β
Simplifying by canceling out like terms:
cos² β(1 - cos² α) = 1/4 + 2cos α sin β
cos² β(sin² α - 1) = -(1/4) + 2sin α sin β
Next, we'll use the Pythagorean identity cos²x + sin²x = 1 to rewrite the equation:
cos² β(sin² α) = 1/4 + 2cos α sin β
sin² β(cos² α) = -(1/4) + 2sin α sin β
Now, we can divide these equations:
cos² β / sin² β = (1/4 + 2cos α sin β) / (-(1/4) + 2sin α sin β)
Simplifying the right-hand side:
(4 + 8cos α sin β) / (-1 + 8sin α sin β) = cos² β / sin² β
Cross-multiplying:
(4 + 8cos α sin β)sin² β = (-1 + 8sin α sin β)cos² β
Expanding both sides:
4sin² β + 8cos α sin α sin β² = -cos² β + 8sin α sin² β
Next, let's use the Pythagorean identity sin²x = 1 - cos²x:
4(1 - cos² β) + 8cos α sin α (1 - cos² β) = -(1 - cos² β) + 8sin α(1 - cos² β)
Simplifying:
4 - 4cos² β + 8cos α sin α - 8cos α sin α cos² β = -1 + cos² β - 8sin α + 8sin α cos² β
Combining like terms:
5 - 4cos² β - 8sin α = 9cos² β + 8sin α
Rearranging the equation:
13cos² β + 8sin α + 4cos² β = 5 + 8sin α
Simplifying:
17cos² β + 8sin α = 5 + 8sin α
Now, let's substitute cos² β with 1 - sin² β:
17(1 - sin² β) + 8sin α = 5 + 8sin α
Expanding:
17 - 17sin² β + 8sin α = 5 + 8sin α
Moving the terms involving sin α to one side:
17sin² β - 8sin α = -12
Dividing both sides by sin β:
17sin β - 8/sin β = -12
Now, we have a quadratic equation:
17sin² β - 12sin β - 8 = 0
Solving this equation will give us the values of sin β.