Question

sin(2Θ-π)= sinΘ
SOLVE ON THE INTERVAL 0≤Θ<2π

Answer 1

@satellite73

Answer 2

i would start with
\[\theta=20-\pi\]

Answer 3

of course \(20-\pi\) is not in that interval so keep subtracting \(2\pi\) until it is

Answer 4

what?

Answer 5

can you post a screen shot of the actual question?

Answer 6

well would you 1st plug in double angle identity on left side

Answer 7

its exactly what ive posted

Answer 8

no, both \(20\) and \(\pi\) are numbers
and so there is no "double angle" identity to use

Answer 9

and sorry i meant sum and difference identity not double angle

Answer 10

its not 20, its 2theta

Answer 11

the only problem is that \(20-\pi\) in the interval \(0<\theta<2\pi\) it is way bigger

Answer 12

ooooh!

Answer 13

\[\sin(2\theta-\pi)=\sin(\theta)\] like that?

Answer 14

yep!

Answer 15

ok now we can get somewhere

Answer 16

ok so whatd be 1st step

Answer 17

i would write
\[\sin(2\theta-\pi)=-\sin(2\theta)\] first

Answer 18

then use the double angle formula

Answer 19

why negative

Answer 20

same reason that
\[\sin(x-\pi)=-\sin(x)\]
if you look on the unit circle you will see why

Answer 21

so 2sinΘcosΘ-1= -sin2Θ ?

Answer 22

then the double angle formula for \(\sin(2\theta)\) gives
\[-2\sin(\theta)\cos(\theta)=\sin(\theta)\]

Answer 23

are u saying it could be either? or was i wrong?

Answer 24

i think i might have lost you somewhere in there
want to go slow?

Answer 25

cuz this is what i got. 2sinΘcosΘ-1= -sin2Θ ?

Answer 26

on the right hand side of the equal sign you have \(\sin(\theta)\) right?

Answer 27

sin2Θ=2sinΘcosΘ thats the formula

Answer 28

& yea i do

Answer 29

so this is the original question \[\sin(2\theta-\pi)=\sin(\theta)\]

Answer 30

but u said to make it -sin2Θ

Answer 31

okay. then?

Answer 32

lets leave the right hand side alone

Answer 33

ok

Answer 34

and work only with
\[\sin(2\theta-\pi)\]

Answer 35

so 1st step?

Answer 36

rewrite \(\sin(2\theta-\pi)\) as \(-\sin(2\theta)\)

Answer 37

so far so good?

Answer 38

ok so π is gone?

Answer 39

yes

Answer 40

ok so that equals right hand side. so far i get it

Answer 41

because it is always the case that
\[\sin(\xi-\pi)=-\sin(\xi)\]

Answer 42

ok

Answer 43

ok now use the "double angle" formula for \(-\sin(2\theta)\)

Answer 44

so itd be same formula only it would hav negative in front?

Answer 45

yes exactly

Answer 46

that is
\[\sin(2\theta)=2\sin(\theta)\cos(\theta)\]
so \[-\sin(2\theta)=-2\sin(\theta)\cos(\theta)\]

Answer 47

-2sinΘcosΘ=cosΘ

Answer 48

yeah, but you messed with the right hand side,
it is still \(\sin(\theta)\)

Answer 49

sorry i meant that

Answer 50

ok good
now set it equal to zero, by adding \(\sin(\theta)\cos(\theta)\) to both sides, giving
\[2\sin(\theta)\cos(\theta)+\sin(\theta)=0\]

Answer 51

ok so bring everything to one side. then what?

Answer 52

factor out the common factor of \(\sin(\theta)\)

Answer 53

when you factor what does it look like

Answer 54

\[\sin(\theta)\left(2\cos(\theta)+1\right)=0\]

Answer 55

and then solve for each part? set each equal to 0 and solve?

Answer 56

set each factor equal to zero, solve for \(\theta\)
that is, solve
\[\sin(\theta)=0\]and solve
\[2\cos(\theta)+1=0\]exactly, what you said

Answer 57

so answers would be π, 2π/3 & 4π/3

Answer 58

thank you!

Answer 59

is \(0\) in your interval?

Answer 60

0 as well

Answer 61

if the interval is \(0\leq \theta\leq 2\pi\) i.e. the closed interval, you have for sine 3 solutions
\[0,\pi,2\pi\]

Answer 62

no not 2π

Answer 63

for cosine you have the right two solutions

Answer 64

cuz it has to be less than 2π

Answer 65

ok
yw