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    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°
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    (Original post by Carlos Nim)
    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°
    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.
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    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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