Question

Use the function defined below to find an equation of the tangent line at the point indicated.

y = cos(x), x = pi/3

Answer 1

I got the slope of \[-\sqrt{3}/2\] I am having trouble making the equation

Answer 2

do you have the point?

Answer 3

x=pi/3
y=cos(pi/3)=1/2

Answer 4

it is
\[(\frac{\pi}{3},\frac{1}{2})\]so point - slope it

Answer 5

m=-(sqrt3)/2

Answer 6

\[(x=\frac{\pi}{3},y=?)\]

Answer 7

plug the values in the formula

Answer 8

I need step by step AFTER you get the slope.

Answer 9

y-(1/2)=(-sqrt3)/2(x-1/2)

Answer 10

uzma is that right?

Answer 11

i think it should be y-y_1=m(x-x_1)

Answer 12

isnt it?

Answer 13

yes right
y1=1/2
x1=pi/3

Answer 14

\[m=-\sqrt{3}/2\]

Answer 15

\[y-\frac{1}{2}=- \frac{\sqrt{3}}{2}(x-?)\]

Answer 16

I'm so lost.

Answer 17

yes it should be x-pi/3

Answer 18

right
he just used point slope formula

Answer 19

\[y-\frac{1}{2}=-\frac{\sqrt{3}}{2}(x-\frac{\pi}{3})\] is my answer

Answer 20

Look, I have the slope. Can somebody PLEASE give me STEP BY STEP explanation of what the heck I need to do.

Answer 21

we agree with you satellite

Answer 22

then with slope either you should have a y intercept or a point on the line

Answer 23

ham
if you plug x into get y there we know a point on the line
that is the point we plugged in into the point slope formula

Answer 24

yes you use
\[y-y_1=m(x-x_1)\] with
\[m=-\frac{\sqrt{3}}{2}, x_1=\frac{\pi}{3}, y_1=\frac{1}{2}\]

Answer 25

my sentences suck

Answer 26

lol

Answer 27

you r given x=pi/2, using this in the curve u can find y which is -(sqrt3)/2

Answer 28

Why didn't you put it into y=mx+b form?

Answer 29

you can

Answer 30

what is b over here?

Answer 31

\[y=\cos(x)=\frac{1}{2} when x=\frac{\pi}{3}\]
\[y=mx+b\]
\[\frac{1}{2}=\frac{-\sqrt{3}}{2}\cdot \frac{\pi}{3}+b\]
solve for b the y-intercept
\[\frac{1}{2}+\frac{\sqrt{3}\pi}{6}=b\]
so we can write:
\[y=\frac{-\sqrt{3}}{2}x+\frac{1}{2}+\frac{\sqrt{3} \pi}{6}\]