Question
a) Find a linear approximation of y=sinx at x=pi/6
b) use part (a) to approximate sin(61pi/360) and sin(59pi/360)
I just really have no idea how to approach this problem. I know the formula is y=f(a)+f'(a)(x-a). Does that mean it would be (sin(pi/6))+(0)((pi/6)-(sin(pi/6))? Any help would be appreciated.
b) use part (a) to approximate sin(61pi/360) and sin(59pi/360)
I just really have no idea how to approach this problem. I know the formula is y=f(a)+f'(a)(x-a). Does that mean it would be (sin(pi/6))+(0)((pi/6)-(sin(pi/6))? Any help would be appreciated.
Answers
"a) Find a linear approximation of y=sinx at x=pi/6 "
no linear approximation needed here, you MUST know that sin(π/6) = 1/2
for b) sin(61π/360)
= sin(60π/360 + π/360)
= sin(π/6 + (π/360)
so we will now use the formula that you stated:
f(x+a)=f(a)+f'(a)(x-a)
recall that if f(x) = sinx, then f '(x) = cosx
so let x = π/6 and a = π/360
sin(61π/360) = sin(π/6) + cos(π/6)*(π/360)
= 1/2 + (√3/2)(π/360)
= .507557
real answer: sin(61π/360) = .507538
error of .000019 , not bad
for the second part, note that 59π/360
= π/6 - π/360
so the last factor would be -π/360
and you should get .49244..
compared to the real answer of .49242..
(again, very close)
no linear approximation needed here, you MUST know that sin(π/6) = 1/2
for b) sin(61π/360)
= sin(60π/360 + π/360)
= sin(π/6 + (π/360)
so we will now use the formula that you stated:
f(x+a)=f(a)+f'(a)(x-a)
recall that if f(x) = sinx, then f '(x) = cosx
so let x = π/6 and a = π/360
sin(61π/360) = sin(π/6) + cos(π/6)*(π/360)
= 1/2 + (√3/2)(π/360)
= .507557
real answer: sin(61π/360) = .507538
error of .000019 , not bad
for the second part, note that 59π/360
= π/6 - π/360
so the last factor would be -π/360
and you should get .49244..
compared to the real answer of .49242..
(again, very close)
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