Question

Is this correct? @vocaloid

Answer 1

good for 14-15

I honestly have no clue how to solve 16 w/o a calculator, I'll try and find something on the web

Answer 2

ok here goes:

we want to re-write the arcsin(x/2) as a sin

so we introduce a new variable (let's call it "a")

such that

(x/2) = sin(a)

and

a = arcsin(x/2) which will replace the expression inside the parentheses

Answer 3

so instead of tan(arcsin(x/2)) we have tan(a) (since we replaced the arcsin with a)

tan(a) = sin(a)/cos(a) just using the definition of tan

cos(a) = sqrt(1-sin^2(a)) (using the cos^2 + sin^2 = 1 identity)

so tan becomes: [let me draw this one]

Answer 4

now, we go back to the top where we said that (x/2) = sin(a) so we replace all the sin(a)'s with x/2

Answer 5

now, we can multiply the numerator and denominator by 2 to get rid of the fraction in the top

Answer 6

then simplify (x/2)^2 to get x^2/4

Answer 7

actually since they might not like having fractions within fractions let's take it a step further

Answer 8

we can change the 1/1 to a 4/4 so that both fractions in the denominator have the same denominator, then combine them to get:

Answer 9

sqrt(1/4) is just 1/2 and since we are dividing by 1/2 that's the same as multiplying by 2

Answer 10

cross out the 2's in the num/denom, then the final sol'n is

Answer 11

x/sqrt(4-x^2) is a bit better than the first sol'n we got since it's a little easier to read

Answer 12

since the arc function and the corresponding non-arc function cancel out then yeah I'd say both are true

Answer 13

Truen, because the arc function and the corresponding non-arc function cancel out

Answer 14

good

Answer 15

idid the first 2

Answer 16

good

for 21 and 22 you just want to solve by factoring

for example 2cos^2(theta) - 3cos(theta) +1 = 0 is like solving

2x^2 - 3x + 1 = 0

Answer 17

when you factor 2x^2 - 3x + 1 = 0

what did you get for x?

Answer 18

1,1/2

Answer 19

good, so we know that cos(theta) = 1 and cos(theta) = 1/2 are the factors, look on the unit circle to find the appropriate theta values

Answer 20

pi/3

Answer 21

no zero and pi/6

Answer 22

pi/6 would be sin(theta) = 1/2 not cos

Answer 23

pi/3 , 0 , 5pi/3

Answer 24

good, repeat the same process for 21

Answer 25

okay so factor it first and then use the unit circle right

Answer 26

yes

they give you sin^2(theta) + cos(theta) = 2

we can replace sin^2(theta) with 1 - cos^2(theta) to get

1 - cos^2(theta)+ cos(theta) = 2

factor and use the unit circle to find theta

Answer 27

1-x^2+x=2

Answer 28

I'm trying to go off of what we did earlier

Answer 29

come to think of it this doesn't factor so hold on

Answer 30

Sure

Answer 31

ok, I just double checked and there's no sol'n so I guess you could write

1 - cos^2(theta)+ cos(theta) = 2 has no real sol'n since this expression is not factorable

Answer 32

If you could check this, I would appreciate it

Answer 33

good

Answer 34

a) if it's been pulled 5 cm we would expect it to oscillate a maximum of 5 cm from the rest position, so 5 cm
b) this is the same thing as period which is already given as pi
c) frequency = 1/period
d) y = Acos(Bx) where A is the max distance from equilibrium (5) and B can be found by solving

period = 2pi/B

Answer 35

A. 5cm
B. 2pi
C. 1/2pi

Answer 36

B) the period is pi not 2pi

Answer 37

Okay and c would be 1/2pi

Answer 38

if the period is pi then 1/period = ?

Answer 39

oh shoot, I'm sorry it would be 1/pi for c
d. y=cos5(bx)

Answer 40

if period = pi = 2pi/B what does B equal?

Answer 41

But i would solve for b so

Answer 42

the 5 needs to be at the front

Answer 43

y = 5cos(Bx)

Answer 44

\[y = 5\cos(2x)\]

Answer 45

good that's it

Answer 46

a. What is the maximum displacement form equilibrium of the object?
5cm
b. What is the time required for one oscillation?
pi
c. What is the frequency?
1/pi
d. Write an equation to model the motion of the object.
y=5cos(2x)

Answer 47

yeah that's good

Answer 48

i've got like 15 minutes before I have to leave for class

Answer 49

okay can we move ot physics

Answer 50

ok