Help me with Calculus

Question

Use an appropriate local linear approximation to estimate the value of the given quantity (cos32°)

Answer 1

I am not 100% sure exactly what is being asked for but you can use differentials to estimate cos(32^o).

d(cos(x)) = -sin(x)dx where dx is approximately delta x which is x₂ - x₁ where x₂ = 32^o, x₁ = 30^o

substituting d(cos(x)) = -sin(30^o)(32^o-30^o) = -sin(30^o)(2^o) but the 2^o must be expressed in radians, so

d(cos(x)) = -sin(30^o)(2pi/180) = -.017453... So, the change in cos(x) from 30^o to 32^o is -.017453...

So the change in cos(32^o) from cos(30^o) will be -.017453...,

Therefore, cos(32^o) is approximately cos(30^o) -.017453 = .848572

The actual value of cos(32^o) = .848048...