d/dxcosx=-sinx

Question

I need a proof using cosx=sinx/tanx.  I have

 

(sinx/tanx)'=quotient rule=(tanx(cosX)-sinx(sec^2x))/ tan^2x

 

= (sine-sinx(sec^2x))/ tan^2x

 

I need the rest from there

Answer 1

Prove d[cos(x)]/dx = -tan(x)  using the identity cos(x) = sin(x)/tan(x).

 

Start with the Quotient Rule: [u/v]′= {v u′ - u v′} / v²

 

d/dx(cosx) = d/dx( sin(x) / tan(x) ) = { tan(x) [sin(x)]′ - sin(x) [tan(x)]′ } / tan²(x)

 

                 = { tan(x) cos(x) - sin(x) sec²(x) } / tan²(x) =

 

                 =  { cos(x) sin(x) / cos(x) - sin(x) sec²(x) } / tan²(x)   ## Simplify tan(x) cos(x) = sin(x)

 

                 = { sin(x) - sin(x) sec²(x) } / tan²(x)   ##  use the Pythagorean identity below

 

                 = { sin(x) [1 - sec²(x)] } / tan²(x)       ## Factor out sin(x)

 

                 = { sin(x) [- tan²(x)] } / tan²(x)       ## note: 1 - sec²(x) = - tan²(x) since 1 + tan²(x) = sec²(x)

 

                 =  - sin(x)   done!!