Question

CHeck please?

Answer 1

@Vocaloid

Answer 2

most likely they want the exact solutions not the approximated forms

from the earlier proof we saw that sin^2(x)cos^2(x) = [1 - cos(4x)]/8

set [1 - cos(4x)]/8 = (2 - sqrt(2))/16 and solve for cos(4x) then find the appropriate angles on the unit circle

Answer 3

pi/16 + pi(n)/2

7pi/16 + pi(n)/2

Answer 4

on the UC this is:

Answer 5

do i need to do anything witht he unit circle or is this the final answer

Answer 6

just a minor correction, it's not pi*n/2 , it's 2*pi*n

so

pi/16 + 2pi(n)

7pi/16 + 2pi(n)

is it

Answer 7

for example, for a) since tan(theta/2) = tan(pi/8), theta must be pi/4, plug in theta = pi/4 into the formula (1 - cos(theta))/sin(theta), and repeat this process for b and c using the appropriate angle values/identity

Answer 8

since cos(theta/2) = cos(pi/8), theta must be pi/4, plug in theta = pi/4 into the formula (sqrt 1+cos(theta) / 2

Answer 9

like this

Answer 10

oh wait you're working on the cos one

Answer 11

yeah that's right, make sure to plug in theta = pi/4 into the formula

Answer 12

since cos(theta/2) = cos(pi/8), theta must be pi/4, plug in theta = pi/4 into the formula (sqrt 1+cos(pi/4) / 2

Answer 13

good, as a minor correction, the square root needs to be on the outside

sqrt[(1+cos(pi/4)/2]

Answer 14

since tan(theta/2) = tan(pi/8), theta must be pi/4, plug in theta = pi/4 into the formula (1 - cos(pi/4))/sin(pi/4)

Answer 15

and then this for the first one

Answer 16

yes

Answer 17

since tan(theta/2) = tan(pi/16), theta must be pi/8, plug in theta = pi/8 into the formula formula (1 - cos(pi/8))/sin(pi/8)

Answer 18

good, that's it for c

Answer 19

start with sin(2 * theta) = 2sin(theta)cos(theta) to re-write the formula in terms of sin only

then plug in h = 100 and v = 60 and solve for theta

Answer 20

the original equation is

h = v^2/16 * sin(theta)cos(theta)

we can use the identity

sin(2 * theta) = 2sin(theta)cos(theta)

to re-write "sin(theta)cos(theta)" as (1/2)sin(2theta)

therefore:


h = (1/2)(v^2/16) * sin(2theta)

solve for theta

Answer 21

yeah come to think of it since sin(2theta) = 200/225 then you pretty much have to use your calculator here

Answer 22

so is that final calculationcorrect for part A

Answer 23

100 = ((1)/(2))((60^(2))/(16)) * sin(2t)
sin(2theta) = 200/225

Answer 24

yeah just be sure to specify that it's in radians

Answer 25

um pi/2

Answer 26

actually hold on a sec

Answer 27

im not sure

Answer 28

actually since it's the product of sin and cos, it takes its maximum at the angle (pi/4), plug in theta = pi/4 into (1/2)(v^2/16) * sin(2theta) to get a maximum of 225/4

Answer 29

so this would be part b right

Answer 30

yes