Question

What is the distance between (0, -sin x) and (-cos x, 0)​

Answer 1

Step-by-step explanation:

Distance:√(0-cosx) ²+(0-sinx) ²

=√(cos²x+sin²x) (sin²x+cos²x=1)

=1

Answer 2

distance between (0, -sin x) and (-cos x, 0)​ = 1

Step-by-step explanation:

the distance between (0, -sin x) and (-cos x, 0)

Distance between Two point  A ( Ax , Ay)   & B(Bx , By)  

= √((Bx - Ax)² + (By - Ay)²)

A = (0, -sin x)  

Ax =  0

Ay = -sinx

B = (-cosx , 0)

Bx = -cosx

By = 0

Distance = √((-cosx - 0)² + (0 -(-sinx))² )

=> Distance = √(cos²x+ sin²x)

cos²x+ sin²x = 1

=> Distance = √1

=> Distance = 1

distance between (0, -sin x) and (-cos x, 0)​ = 1

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