Question

Simplify trig problem?

Answer 1

\[\sqrt{(1-\cos \theta})(1+\cos \theta)\]

@phi

Answer 2

that square root goes over whole equation

Answer 3

multiply it out. what do you get ?

Answer 4

it wont insert but |sin thete|?

Answer 5

theta

Answer 6

you should get
\[ 1 - \cos^2 \theta\]
and that simplifies to sin^2
then take the square root.

Answer 7

sin(theta)

Answer 8

yes, unless we want complex numbers, we should put absolute value signs around sin theta

Answer 9

I was trying to do that but it won't post, that why I had this -->
\left| sin Theta \right|

Answer 10

okay thanks can you help with one more?

Answer 11

The equation editor probably can do it for you

Answer 12

Which function has the largest maximum?
f(x)
g(x)
h(x)
All three functions have the same maximum value.

Answer 13

g(x) = 4cos(2x-pi) -2
h(x) = -(x-5)^2 +3

this one is a table
f(x) =
x, 0 1 2 3 4 5 6
y, -5 0 3 4 3 0 -5

Answer 14

i'm not sure how you find the maximum

Answer 15

what is the max of f(x) , based on the table?

Answer 16

the max number? 6

Answer 17

f(x) "takes an x" and returns a y
we want the max y value

Answer 18

4

Answer 19

ok , max for f(x) is 4

now for h(x)
h(x) = -(x-5)^2 +3
there are two parts to h(x): - (x-5)^2 and +3
first look at (x-5)^2 , can that ever be negative ?

Answer 20

no it will stay positive

Answer 21

to figure out (x-5) * (x-5)
we have these choices: x-5 is negative. negative times negative is positive
x-5 is 0, 0* 0 is 0
(x-5) is positive, positive * positive is positive
so 0 or positive

Answer 22

but we have -(x-5)^2 and that minus sign means we will get 0 or negative.

Answer 23

so h(x) = ( 0 or negative) + 3
what is the max value we could get out of that ?

Answer 24

3

Answer 25

now g(x)

g(x) = 4cos(2x-pi) -2

cos(angle) has a max value of what ?

Answer 26

cosine wanders between what and what y values?

Answer 27

2? im not sure on this one

Answer 28

oh. cosine looks like this