`cos(90^(@)30')`, approximately given that `1^(@)=0.0175` is

Options

A. `-0.0082`
B. `-0.0087`
C. `0.0087`
D. `0.0081`

Answer

B

Explanation

Let `f(x)=cos x`
`:.f'(x)=-sinx`
take `a=90^(@)=(pi)/2`
then `f(a)=f(90^(@))=cos90^(@)=0`
`f'(a)=f'(90^(@))=-sin90^(@)=-1`
Now, `h=30'=30/60xx1^(@)=3/6xx0.0175=0.0087`
`:.f(a+h)=f(a)+h.f'(a)`
`:.cos(90^(@)30')=0+(0.0087)xx(-1)=-0.0087`