Question

Cos theta minus alpha equals to b sin theta + beta equals to q then prove p square + q square minus 2 pq sin alpha + beta = 2 cos square alpha + beta

Answer 1

The correct equation is: p square + q square minus 2 pq sin alpha + beta =  cos square alpha + beta

Given:

Cos theta minus alpha equals to b sin theta + beta equals to q

cos (θ - α) = p, sin (θ + β) = q .......(1)

To prove:

p square + q square minus 2 pq sin alpha + beta =  cos square alpha + beta

p² + q² - 2 pq sin (α + β) = cos² (α + β)

Proof:

Let us consider,

α + β =  (θ + β) - (θ - α)

taking sin on both the sides, we get,

sin (α + β) =  sin [ (θ + β) - (θ - α) ]

sin (α + β) = sin (θ + β) cos (θ - α) - cos (θ + β) sin (θ - α)

sin (α + β) = qp - √(1 - q²) √(1 - p²) .......(2)

w.k.t cos² x = 1 - sin² x, so we get,

cos²  (α + β) = 1 - sin² (α + β)

∴ cos²  (α + β) = 1 - [ qp - √(1 - q²) √(1 - p²) ]²

further solving, we get,

cos²  (α + β) = p² + q² - 2 p²q² + 2 pq [ √(1 - q²) √(1 - p²) ] .........(3)

Now consider,

p² + q² - 2 pq sin (α + β)

= p² + q² - 2 pq [ qp - √(1 - q²) √(1 - p²) ]

= p² + q² - 2 p²q² + 2 pq [ √(1 - q²) √(1 - p²) ]

∴ p² + q² - 2 pq sin (α + β) = p² + q² - 2 p²q² + 2 pq [ √(1 - q²) √(1 - p²) ]

using (3), we get,

p² + q² - 2 pq sin (α + β) = cos²  (α + β)

Hence proved.