Question

The distance between the points ( cosA, sinA ) and ( -sinA, cosA ) is

Answer 1

Hi...☺

Here is your answer...✌
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Let the points be A (cosA,sinA) and
B (-sinA,cosA)

Using Distance formula,

We have,

$AB = \sqrt{ ( - \sin(a) - \cos(a)) {}^{2} + ( (\cos(a) - \sin(a)) {}^{2} } \\ \\ = \sqrt{ \sin {}^{2} (a) + \cos {}^{2} (a) + 2 \cos( a) \sin(a) + \cos {}^{2} (a) + \sin {}^{2} (a) - 2 \sin(a) \cos(a) } \\ \\ = \sqrt{1 + 1} \\ \\ = \sqrt{2}$

Answer 2

√2 is The distance between the points ( cosA, sinA ) and ( -sinA, cosA )

Step-by-step explanation:

we need to find The distance between the points ( cosA, sinA ) and ( -sinA, cosA )

Distance between  Two Points  A (x₁ , y₁) & B (x₂ , y₂)

is given by √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = CosA  y₁ = SinA

       x₂ = -SinA  y₂ = CosA

Distance = √(-SinA - CosA)² + (CosA - SinA)²

=√Sin²A + Cos²A + 2SinACosA  + Cos²A + Sin²A - 2SinACosA

as we know that Sin²A + Cos²A = 1

= √1 + 2SinACosA  +1 - 2SinACosA

= √2

The distance between the points ( cosA, sinA ) and ( -sinA, cosA ) is √2

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