#Trig equation
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So I know that the amplitude is 6
and the axis of curve is y=10
and the k value is 30
If by that you mean the average value, then no.
but im not sure how to solve for phase shift
What is k?
it literally is tho
16+4/2 = 10
so y=10
Oh, yeah, sorry. I misread a bit.
its alr
Oh. A bit of an odd form. I'm more used to f(x) = A cos(ωx + φ0) + B.
interesting
In any case, your k is my ω.
is that the standard form in textbooks?
In which case ω should be 2π/12 = π/6, rather.
It's the standard form used in physics.
Doesn't matter.
ok
The only thing that will differ is the initial phase.
so so far i have f(x)=6sin(30(x-c))+10
Again, doesn't seem right. Why 30?
because k=360/12
The relation connecting the period and frequency is ω = 2π/T.
Ohh, you meant 30°, not just 30.
yes 30 degrees
If so, yeah, k = 30°.
my bad
So, we have:
f(x) = 10 + 6sin(30°(x - d))
We know that f(3) = 4. So, try using that to find d.
hmm okay so i sub in the point?
Yeah.
Well, you do need to remember how to solve elementary trigonometric equations.
Bring the equation to the elementary form first.
Elementary?
Like the 10 in front?
Elementary trigonometric equations have a form f(x) = a, where the f(x) is a trigonometric function.
So, isolate the trigonometric function first.
So divide 6sin?
Well, bring all the constant terms to the right first. Then yes.
Yeah idk what I’m doing 💔💔
We have:
10 + 6sin(30°(3 - d)) = 4
6sin(30°(3 - d)) = -6
sin(30°(3 - d)) = -1
This is now an elementary trigonometric equation. So, recall how to solve it.
Would you do inverse sine
Which gives -90=(30(3-d)
Then divide by 30
3-d=-3
d=6?
That is one solution.
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