The 3rd degree taylor polynomial for cos(x) centered at a = π 2 is given by, cos(x) = − x − π 2 + 1 6 x − π 2 3 + r3(x). using this, estimate cos(88°) correct to five decimal places.

0.034899

Step-by-step explanation:

There are quite a few mistakes in writing the the question, but i corrected them below. firstly the value of a that you stated seems incorrect and based on the last part of the question asking for cos(88°) i figured that what you were trying to write down was a=π/2 , since 88° is in the vicinity of a=π/2 or 90° and thats what the taylor series expansions are all about, that is estimating the values of functions about a given point. The second problem was in writing down the taylor series itself. So i used the formula for the taylor series expansion and obtained the correct version about the point a=π/2 that is,

In the last step we converted degrees into radians and then plugged it in.

If you use the calculator to evaluate cos(88°) you would get the same result to 5 decimal places.

ce pula mea vrei ma

step-by-step explanation:

gayule

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12