- #1
supernova1203
- 210
- 0
I understand how he got -120 and 360 as zeros, but then there is another zero 240, how did he get that?
Check the diagram i have posted (which was done by the teacher) for simpler version
A bit of background:
we are working with sine functions the equation being
y=2sin(x-30)+1
so we re arrange y=sin(x-30)=-1/2
using special triangles/angles learned in previous lessons i deduce that since we are working with with sin and special angle 30, the ratio is 1/2 but since it is negative i also deduce that using the CAST rule (AllA positive in quadrent I ,only sinS positive in quadrent II,
only tan T positive in quadrent III, and only cosC positive in Quadrent IV
hence CAST) that the triangle ( or terminal arms) are in Q III and Q IV
according to the rules i learned here on these very forums
If theta is in QII then 180-theta
If theta is in QIII then theta -180
If theta is in QIV then 360-theta
There is no referrence angle in QI because in QI the main angle is the same thing as the referrence angle
using those rules i find QIII = theta - 180
30-180 = -150
then sub it back into the original equation sin(x-30)=-1/2
x-30=-150
x=-150+30
x=-120
One of the zeros is -120
similarly
QIV = 360-theta
360-30
=330
sub into original equation
sin(x-30)=-1/2
x-30=330
x=330+30
x=360
One of the zeros is 360...
the teacher somehow finds another zero which is 240...i have no idea how.Ok i see how he did it, but it makes no sense when i try to use the rules
"If theta is in QII then 180-theta
If theta is in QIII then theta -180
If theta is in QIV then 360-theta
There is no referrence angle in QI because in QI the main angle is the same thing as the referrence angle"
because he is doing it visually, and I am doing it the non visual way
Check the diagram i have posted (which was done by the teacher) for simpler version
A bit of background:
we are working with sine functions the equation being
y=2sin(x-30)+1
so we re arrange y=sin(x-30)=-1/2
using special triangles/angles learned in previous lessons i deduce that since we are working with with sin and special angle 30, the ratio is 1/2 but since it is negative i also deduce that using the CAST rule (AllA positive in quadrent I ,only sinS positive in quadrent II,
only tan T positive in quadrent III, and only cosC positive in Quadrent IV
hence CAST) that the triangle ( or terminal arms) are in Q III and Q IV
according to the rules i learned here on these very forums
If theta is in QII then 180-theta
If theta is in QIII then theta -180
If theta is in QIV then 360-theta
There is no referrence angle in QI because in QI the main angle is the same thing as the referrence angle
using those rules i find QIII = theta - 180
30-180 = -150
then sub it back into the original equation sin(x-30)=-1/2
x-30=-150
x=-150+30
x=-120
One of the zeros is -120
similarly
QIV = 360-theta
360-30
=330
sub into original equation
sin(x-30)=-1/2
x-30=330
x=330+30
x=360
One of the zeros is 360...
the teacher somehow finds another zero which is 240...i have no idea how.Ok i see how he did it, but it makes no sense when i try to use the rules
"If theta is in QII then 180-theta
If theta is in QIII then theta -180
If theta is in QIV then 360-theta
There is no referrence angle in QI because in QI the main angle is the same thing as the referrence angle"
because he is doing it visually, and I am doing it the non visual way