Question

simplify. sin(3pi/2 + theta)

Answer 1

@satellite73 please help me

Answer 2

have you tried using:
\[\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\]

Answer 3

yea but how do I use that?

Answer 4

simplify???

Answer 5

well sin(3pi/2+theta)=sin(a+b)
if choose a to be 3pi/2 and b to be theta

Answer 6

so replace the a's with 3pi/2 and the b's with theta's

Answer 7

I'm stuck here:
\[\cos 3\pi/2 * \cos x + \sin 3\pi/2 * \sin x\]

Answer 8

well first of all you did cos(3pi/2-theta)
instead of sin(3pi/2+theta)

Answer 9

well and I guess your x is theta

Answer 10

wait a sec why are we using sin(a+b)??

Answer 11

is this not your question "simplify. sin(3pi/2 + theta)
"

Answer 12

do you see the sin in front?

Answer 13

oh wait nvm, I was lookign at the wrong question disregard everything I said, carry on with the question posted

Answer 14

yea sorry keep going

Answer 15

let me know when you need more help

Answer 16

ok I'm stuck here:
\[\sin 3\pi/2 * \cos \theta + \cos 3\pi/2 * \sin \theta \]

@freckles

Answer 17

ok
cool
do you know what sin(3pi/2)=?
and what cos(3pi/2)=?
use unit circle if you don't know off the top of your head

Answer 18

how do I use the unit circle for this?

Answer 19

I do not remember my unit circle values, but I'mpretty familiar with it

Answer 20

http://etc.usf.edu/clipart/43200/43217/unit-circle9_43217_md.gif

do you see anywhere on this unit circle (This picture) 3pi/2?

Answer 21

270 degrees

Answer 22

notice the coordinates at that position

Answer 23

(x,y)
=(cos(theta),sin(theta))

Answer 24

this means the x value is telling you what the cosine value is there
and the y value is telling you what the sine value is there

Answer 25

oh, so should I substitute?

Answer 26

can you tell me now what sin(3pi/2)=?
and what cos(3pi/2)=?

Answer 27

yea just a sec. im going to write it real quick

Answer 28

now I'm left at
\[-\sin 3\pi/2\]

Answer 29

where did you get that from?

Answer 30

are you still trying to evaluate cos(3pi/2) and sin(3pi/2)

Answer 31

yes

Answer 32

do you see at 3pi/2 the coordinates there are (0,-1)?

Answer 33

do you know the first value is the cosine of 3pi/2
and the second value is the sin of 3pi/2

Answer 34

\[(\sin (3\pi/2) * 0) + (\sin (3\pi/2) * -1)\]

Answer 35

I don't get what you are doing

Answer 36

what happen to your thetas

Answer 37

so

0 + (-sin(3pi/2))

Answer 38

isnt it (x,y) = (cos(theta), sin(theta))

Answer 39

\[\sin 3\pi/2 * \cos \theta + \cos 3\pi/2 * \sin \theta \]
you said you got here
and I'm telling you the cosine of 3pi/2 is 0
and the sine of 3pi/2 is -1
so just replace sin(3pi/2) with -1 and replace cos(3pi/2) with 0
don't make the thetas just disappear

Answer 40

the x,y are 0,-1 from 270 degrees

Answer 41

oooh, so the sin theta and anything with a theta doesnt disappear

Answer 42

how could they disappear we don't know what theta is

Answer 43

so now im stuck here, feel like I did something wrong:
\[-\cos \theta + 0\]

Answer 44

so \[-\cos \theta \]

Answer 45

yep \[\sin(\frac{3\pi}{2}+\theta) \\ \sin(\frac{3\pi}{2})\cos(\theta)+\cos(\frac{3\pi}{2})\sin(\theta) \\ -1 \cdot \cos(\theta)+0 \cdot \sin(\theta) \\ -1 \cos(\theta)+0 \\ -\cos(\theta)\]

Answer 46

- cos theta is the answer?

Answer 47

I thought it said simplify sin(3pi/2+theta)
I thought we just did that

Answer 48

was there another question

Answer 49

yea can wedo another harder one

Answer 50

if you want to do it
if you need help just post here

Answer 51

\[(\sin x + \cos x)^2 = 1 + \sin(2x)\]

Answer 52

verify trig identity

Answer 53

have you tried expanding (sin(x)+cos(x))^2
recall (a+b)^2=(a+b)(a+b)=a^2+2ab+b^2

Answer 54

heres where i got now:
\[sinx^2 + 2sinxcosx + cosx^2 = 1 + \sin(2x)\]

Answer 55

ok and you know sin^2(x)+cos^2(x)=?

Answer 56

and you also know 2sin(x)cos(x)=?

Answer 57

think Pythagorean identity and double angle identity for sin

Answer 58

umm, wts the pythagorean identity?

Answer 59

recall the unit circle equation is x^2+y^2=1
and here it what it looks like:

The center is (0,0)
and the radius is 1
Say (x,y) falls on the circle and say we put it here:


Well by Pythagorean Theorem we have the following relationship
x^2+y^2=1 which was our equation above
Notice also cos(theta)=x and sin(theta)=x
so we have
\[(\cos(\theta))^2+(\sin(\theta))^2=1 \\ \cos^2(\theta)+\sin^2(\theta)=1\]
this is one of the Pythagorean identities or one of the forms it can take anyways