Question

the maximum value of 4sinx - 3 cos x , on [ 0 , 2pi] , WITHOUT A CALCULATOR

Answer 1

\[f'(x)=4\cos(x)+3\sin(x)=0 \text{ solve for x}\]

Answer 2

ok i got tan x = -3/4

Answer 3

errr, tan x = -4/3

Answer 4

right

Answer 5

arctan (-4/3) = x , now i cant use calculator

Answer 6

lagrange, did it work ?

Answer 7

myin, WHAT ARE THE SOLUTIONS, max and min

Answer 8

the exact x=arctan(-4/3)

Answer 9

now you need to find out if it is a max or min

Answer 10

there is also another x=arctan(-4/3)+180

Answer 11

i will show you how to plug in one of these
say we look at x=arctan(-4/3)+180
and we want to evaluate our function f at this x
we would get
f(arctan(-4/3)+180)=4sin(arctan(-4/3)+180) - 3 cos(arctan(-4/3)+180)
\[=4[\sin(\arctan(-4/3))\cos(180)+\cos(\arctan(-4/3))\sin(180)] \]
\[-3[cos(arctan(-4/3))cos(180)-sin(arctan(-4/3))sin(180)] \]
\[=4[\sin(\arctan(-4/3)(-1)+\cos(\arctan(-4/3))(0)]\]
\[-3[\cos(\arctan(-4/3)(-1)-\sin(\arctan(-4/3))(0)]\]
\[=-4\sin(\arctan(-4/3))+3 \cos(\arctan(-4/3))\]
so now we need to evaluate sin(arctan(-4/3)) and cos(arctan(-4/3))
remember if y=arctan(-4/3) => tan(y)=-4/3

we can find the hyp. by using Pythagorean thm and we get 5 for the hyp
so
sin(y)=-4/5
cos(y)=3/5
so continuing the evaluation we get
\[=-4(-4/5)+3(3/5)=\frac{16+9}{5}=\frac{25}{5}=5\]

Answer 12

now you should evaluate f(arctan(-4/3))

Answer 13

and don't forget to plug in the endpoints

Answer 14

ok thanks so much

Answer 15

wanna fight again?

Answer 16

did you understand what i did above
in that triangle that angle was suppose to be labeled y

Answer 17

oops +pi
not 180

Answer 18

\[[0,2\pi]\]

Answer 19

satellite makes everything such a big deal

Answer 20

he is saying that they want theta in terms of radians

Answer 21

so instead of saying +180
say +pi

Answer 22

no i am saying they want x in terms of numbers

Answer 23

i mean x in terms of radians

Answer 24

and in any case if you are using sine and cosine as a function of degrees, then the derivative of sine is not ( i repeat not) cosine!

Answer 25

x is neither a degree nor a radian nor any other specific unit. x is a number as in
\[f(x)=x^2\]

Answer 26

satellite
an angle is measured in degrees or radians
why do you say these aren't numbers?

Answer 27

30 degrees is not a number?

Answer 28

pi/6 is not a number?

Answer 29

in any case myininays is of course right, just forget the damned degrees unless you are solving a triangle using degrees as your angle measure

Answer 30

i did not say 30 is not a number. what i am trying to say is that sine and cosine are functions of real numbers, and as such, they only correspond to the functions of angles if the angles are measured in radians.

Answer 31

why can't the angles be measured in degrees?

Answer 32

fine, but then the functions are functions of angles measured in degrees, and do not correspond to the functions as functions of numbers. so in particular the derivative of sine is not cosine if you are thinking of these as functions with domain angles measured in degrees rather than numbers.

Answer 33

then what is derivative of sine?

Answer 34

if the angle is measured in degrees?

Answer 35

you can see this clearly from the graph. graph sine and you will see that the x - axis is labeled with numbers like pi, 2pi etc. they are real valued functions with domain all real numbers and period 2 pi

Answer 36

jesus i worked it out once and it is a pain. you have to use the chain rule.

Answer 37

lol ok satellite

Answer 38

http://answers.yahoo.com/question/index?qid=20080129223349AAi6win

Answer 39

yes that makes sense

Answer 40

whew.

Answer 41

i will be quiet now (until the next time...)

Answer 42

i think perl said he gots it from where i left off
right?

Answer 43

actually, not th that you asked, but we can do this problem without calc

Answer 44

satellite never likes mine ways

Answer 45

you can rewrite this as
\[5
\sin(x-\tan^{-1}(\frac{3}{4}))\] which obviously has a max of 5.

Answer 46

just thought i would mentions it. only way i know this is that someone here asked how to write
\[a\cos(x)+b\sin(x)\] as a single trig function and looked it up day before yesterday

Answer 47

\[\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos(x)+\frac{b}{\sqrt{a^2+b^2}}\sin(x))\]
\[\sqrt{a^2+b^2}(\sin(C+x)), \text{where } C=\tan^{-1}(\frac{a}{b})\]

Answer 48

\[4\sin(x)-3\cos(x)=\sqrt{(4)^2+(-3)^2}\sin(\tan^{-1}(\frac{-3}{4})+x)\]

Answer 49

Attached is a plot from x=0 to x=2 pi + 1

Answer 50

remember tan(x) is an add function
so tan(-x)=-tan(x)

so if tan(-x)=y then -tan(x)=y
=> arctan(y)=-x , arctan(-y)=x
=> -arctan(y)=x
so you could write the expression like satellite has

Answer 51

add function lol

Answer 52

oops

Answer 53

just teasing i know what you meant. did you go trick or treating?

Answer 54

no
you knew what i meant after i said tan(-x)=-tan(x) right? :)

Answer 55

yes.
why not? did kids come to the house?