Question

plz show me how to solve this

Answer 1

What is that cosine? It looks like cos(4*2^0)?

Answer 2

cos 42 thats jus the degree sign

Answer 3

Ah, okay. Give me a sec to write it out and I'll post the steps to solve the problem.

Answer 4

thanx alot....take ur time :D:D

Answer 5

Okay, so a Taylor series is a way to express the value of a function is an infinite sum. The Taylor series of a function f(x) about a point a is:\[\sum_{n=0}^{\infty} (f^{(n)}(a)/ n!)(x-a)^n\]where \[f^{(n)}(a)\]is the nth derivative of f evaluated at a.

So for a cosine function about pi/4, this will be

\[cosx = \cos(\pi/4)/0! - \sin(\pi/4)/1!(x-\pi/4) - \cos(\pi/4)/2!(x-\pi/4)^2 + \sin(\pi/4)/3!(x-\pi/4)^3 + ...\]Now cos(pi/4) = -cos(pi/4) = 0, so we can rewrite this as\[cosx = -\sin(\pi/4)(x-\pi/4) + \sin(\pi/4)/6(x-\pi/4)^3 - ...\]or\[cosx = \sum_{n=1}^{\infty}(-1)^{n}\sin(\pi/4)(x-\pi/4)^{n}/(2n-1)!\]

Now it should be clear that the sign of each term is opposite the sign of the previous term, and that the magnitude of each term is smaller than the last. Thus, The total amount of error in the problem is necessarily at most the value of the first term left out (ex, if you approximate cosx by using the first 4 terms of the above series, the maximum possible error left in the expression is equal to the value of the 5th term.) Thus, to find an approximation for cosine to within an error of 10^-6, you must simply find the first term whose magnitude is less than 10^-6 and approximate the cosine using all of the proceeding terms.

(Note that you're going to have to convert x from 42 degrees to radians before plugging it into the expression)

Answer 6

*as an infinite sum

Answer 7

*preceding terms

Answer 8

thanku very much that helpd alot :D

Answer 9

No problem, if you have any other questions post and I'll try to respond.