Sine and cosine functions Zeros

[supernova1203]

How Do i find the zeros for this particular sine function?

y=asin(k(x-d))+c

when we are given a K value i know how to find the zero but when we are not, i don't know how to find the zero(Atleast after everything has been horizontally, and vertically shifted and streched)

y=2sin(x-30 degrees) +1

I have graphed all the points i just don't know how to find the zero here, if i were given

y=2sin3(x+45 degrees) + 2 I would know how to find the zero (I would just come up with my own little equation ) 2sin(3 x what value = 0) but for y=2sin(x-30 degrees) + 1 i just don't know how to find the zeros(the final zeros after everything is where it should be, shifted/streched/transformed etc)

 

[tiny-tim]

hi supernova1203!

(have a degree: ° )

y=2sin(x-30 degrees) +1

I have graphed all the points i just don't know how to find the zero here …

the zeros will be at 2sin(x-30°) +1 = 0,

so the first thing to do is to rewrite it as:

sin(x-30°) = -1/2​

you know that sin(-30°) = -1/2, so obviously one solution will be x-30° = -30°,

and then you can list all the other solutions: x-30° = … ?

 

[LCKurtz]

Say you have y = a*sin(bx + c) + d. To find its zeros you set it equal to 0:

a*sin(bx + c) + d = 0

sin(bx + c) = -d/a.

Now, of course, it the right side is not in the interval [-1,1] there will be no solution. If it is in that interval, then there will be a value for bx + c between -pi/2 and pi/2. Let's call that value θ₀. So you have bx + c = θ₀, but also any angle coterminal with θ₀ will work, so you can have:

(*) bx + c = θ₀ + 2npi

If -d/a is +1 or -1, these values of are all that will work and you can solve for x and you are done.

But usually -d/a is in (-1,1), θ₀ is in quadrant 1 or 4, and there is another quadrant where bx+c can be.

For example, if θ₀ is in quadrant 1, then any angle coterminal with pi - θ₀ will work:

θ = pi - θ₀ + 2npi = (2n + 1)pi - θ₀

If θ₀ is in quadrant 4, then any angle coterminal with pi - θ₀ will work:

θ = pi - θ₀ +2npi = (2n+1)pi - θ₀

As luck would have it, these are the same formula so you can just set

bx + c = (2n+1)pi - θ₀

and solve for x.

 

[supernova1203]

why oh why do they not teach us in radians and with pies?! They use degrees and everywhere I am looking online to find help they use radians and pies...I think they teach us intentionally with degree only so as to handicap us so we can rely solely on the teachers and not get help anywhere else @_@