Question

Use Linear approximation to estimate the following quantities. Choose a value of a to produce a small errror.
\[\cos31 \]
Linear approx form
\[\ L(x) = f(a)+f \prime(a)(x-a) \]

Answer 1

My issue is figuring out how to get the answer using trig functions. According to my book, as I am evaluating my values using the linear approx. formula, my answers have to be in radians. So I want to choose 30 as my a,

\[\cos\frac{ pi }{ 6 }=.866\]
^ This is my value that approximating
Now plugging in pi/6 in the formula

\[\cos\frac{ pi }{ 6 }+(-sin\frac{ pi }{ 6 })(31-30)\]I get .366. Where is my mistake?

Answer 2

I don't think you should mix in degrees and radians.
in other words, change 31-30 = 1deg into radians = \( \frac{\pi}{180} \)

Answer 3

Ah, I see now i got the right answer. Thanks.
So just to make sure i'm doing this right, if i have say tan3, and an a value of 2...
\[\tan3 = tanpi/60=.05\]
\[\tan(\frac{\pi}{90})+\sec^2(\frac{\pi}{90})(\frac{\pi}{180})=.035=.04\]

Answer 4

if you have tan 3º as your starting point , your formula should be
\[ f(x) \approx \tan 3º + sec^2( 3º )(x - 3) \cdot \frac{\pi}{180}\]
so you should get minus pi/180 (not plus pi/180)
in other words
\[ \tan(\frac{\pi}{90})-\sec^2(\frac{\pi}{90})(\frac{\pi}{180}) \]

Answer 5

my a is , 3 is the x value

Answer 6

*a is 2.

Answer 7

if a is 2 then you want
tan(x) = tan(2º) + sec^2(2º) (x - 2º) * pi/180

Answer 8

so your equation looks ok
(but I would use more decimals)

Answer 9

oh ok

Answer 10

thanks.

Answer 11

tan(3º) = 0.052407779

the approximation is
\[ \tan(3º) \approx 0.034920769 + 1.00121946 \cdot \frac{\pi}{180} \\
= 0.034920769 + 1.00121946 \cdot 0.017453292
\\= 0.052395
\]
which is close to the actual value 0.052407779