Find the equation of the tangent at the origin for : siny = 2sinx

Question

Find the equation of the tangent at the origin

Answer 1

(0,0) satisfies the given equation siny = 2sinx

 

using implicit differentiation:  y' cosy = 2cosx so y'(0,0) = 2 since both cos y & cos x are 1.

 

y = 2x is therefore the equation of the tangent at (0,0), using point slope form.

Answer 2

Try implicit differentiation.

 

siny = 2sinx

 

(cosy)y' = 2cosx

 

y' = 2cosx / cosy

 

At the origin, x = y = 0

 

cosx = cos(0) = 1

cosy = cos(0) = 1

 

y'(0,0) = 2(1)/(1) = 2

 

y = mx + b

 

m = y' = 2

 

y = 2x + b

 

b = 0 (goes through origin)

 

y = 2x