IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ssbrightstar1

Find cos 61⁰, it is being given that sin 60⁰ =

0.86603

and1⁰=0.01745 radian?

sagarvaze

cos(60+1) = cos60cos1 - sin60sin1

find cos60 by cos@ = root(1-sin^2@)

for small angle sin@ = @ and cos@ = 1

nknikhilesh1

please make this by error and approximation method.that was the question of ss.

shinee

here let us assume that f(x) is cos60 and dx is 1degree
error f'(x)dx =-sin60X(0.01745)
add to cos60
so, we get 0.5-root3/2(0.01745)
simplify

srujana

From the definition of a derivative,

$f'(x)=\frac{f(x+h)-f(x)}{h}$

$\Rightarrow$

we need to find the value of $cos\ 61^0$

let f(x)=cos x

$f(60^0+1^0)=cos\ 61^0 =f'(60^0).(1^0)+cos\ 60^0$

$=-sin\ 60^0\times (1^0)+cos\ 60^0$

$=-0.86603\times (0.01745)+\sqrt{1-sin^2 60^0}$