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1. 1 ° = 0.0175 rad
Sin 60 ° = 0.8660
Sin 45 ° = 0.7071

f(A) = Cos A
Need to find approximate value for Cos 31 ° and Cos59°

i have no idea how to even start this question....
my only clue so far is that f'(A)= -Sin(A)
and that Sin 60°-Sin 45° = Sin 15°
Sin15° + SIn15° + Sin1°= Sin 31°
Sin 60°-Sin1°=Sin59°
2. (Original post by Carlos Nim)
Sin 60 ° = 0.8660
Sin 45 ° = 0.7071

f(A) = Cos A
Need to find approximate value for Cos 31 ° and Cos59°

i have no idea how to even start this question....
my only clue so far is that f'(A)= -Sin(A)
and that Sin 60°-Sin 45° = Sin 15°
Sin15° + SIn15° + Sin1°= Sin 31°
Sin 60°-Sin1°=Sin59°
I would suggest using the standard identities for cos(A + B) and cos(A - B) applied to cos(30 + 1) and so on (but in radians not degrees), combined with the small angle approximations [email protected] = @ and [email protected] = 1.
3. ok what i did for Cos31°0.305
i don't know Sin 31° but i can calculate Sin° 30 which is 2(0.8660 - 0.7071) = 0.3058

y=Cos A , if A increases , y changes too

Based on that, i have to calculate the increase in y from Sin30° to that of sin31°
Delta y/ Delta x = dy/dx
Delta y = dy/dx. delta x
delta y = -Sin 30°. Sin 1 ° = -0.0053515
therefore Sin31° = 0.3058 + ( -0.0053515 ) = 0.3004485

to be honest, i have no idea what i just did

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