Question

Find the distance between the points (cosb, -sinb) and (sinb, cosb)

Answer 1

Answer:

points are P (cos x, -sinx) and Q (sin x, cos x)

d= √(sin x-cos x)² + (cos x+ sin x)²

d=>√(sin²x+cos²x-2sinxcosx+cos²x+sin²x+2sinxcosx)

d = √(1 + 1) [ 2sinxcosx cancels out ]

d = √2 [ sin²x+cos²x=1 ]

therefore distance = √2 units

Answer 2

Step-by-step explanation:

thats it................snzjjsbshz.....