Find the unknown values in the problem involving trigonometry graphs

[chwala]

Homework Statement

See attached.

Relevant Equations

trigonometry

This is the question...

My attempt on part (i),
$b=\dfrac {16π}{2π}=8$
$11=a sin 32π+c$
$c=11$
$5=-a\frac {\sqrt 3}{2} +11$
$10=-a\sqrt 3+22$
$12=a\sqrt 3$
$a=\dfrac {12}{\sqrt 3}$

Is this correct? Thanks...

 

[Mark44]

Homework Statement:: See attached.
Relevant Equations:: trigonometry

This is the question...

View attachment 300644
My attempt on part (i),
$b$=$\dfrac {16π}{2π}$=$8$
$11=a sin 32π+c$
$c=11$
$5$=$-a$$\frac {\sqrt 3}{2}$ +$11$
$10$=$-a$$\sqrt 3$+$22$
$12=a\sqrt 3$
$a$=$\dfrac {12}{\sqrt 3}$

Is this correct? Thanks...

No. Instead of b = 8, I get b = 1/8, and I get different values for a and c. Since the period is $16\pi$, it must be the case that the coefficient of x is less than 1, not greater than 1. Apparently you didn't check your answers to make sure that you had the correct parameters.

 

[chwala]

No. Instead of b = 8, I get b = 1/8, and I get different values for a and c. Since the period is $16\pi$, it must be the case that the coefficient of x is less than 1, not greater than 1. Apparently you didn't check your answers to make sure that you had the correct parameters.

...sine $90^0=1$ and $sine 30^0=0.5$
ok, i think it ought to be, ...
$11=a+c$
$10=-a+c$

giving us,
$a=0.5$ and $c=10.5$

 

[Mark44]

ok, i think it ought to be, ...
$11=a+c$
$10=-a+c$

giving us,
$a=0.5$ and $c=10.5$

I don't get either of those values. Are you checking them by substituting the given points into the equation?

 

[chwala]

Initially i had,
$11$=$a sin 8(4π)$+$c$
$5$=$a sin 8$$\left[-\frac {4π}{3}\right]$+$c$

which gives me,
$11$=$a sin 32π$+$c$
$5$=$-a sin$$\left[\dfrac {32π}{3}\right]$+$c$

and from the first equation, $sin 32π=0$, and $sin$$\dfrac {32π}{3}$=$\dfrac {\sqrt 3}{2}$

Are you certain that my earlier attempt is not correct?

 

[chwala]

No. Instead of b = 8, I get b = 1/8, and I get different values for a and c. Since the period is $16\pi$, it must be the case that the coefficient of x is less than 1, not greater than 1. Apparently you didn't check your answers to make sure that you had the correct parameters.

True, I see my mistake on $b$...let me look at it again...thanks @Mark44

 

[Mark44]

Initially i had,
$11$=$a sin 8(4π)$+$c$
$5$=$a sin 8$$\left[-\frac {4π}{3}\right]$+$c$

The graph of $y = a\sin(8x) + c$ has a period of $\frac{2\pi}8 = \frac \pi 4$.
This conflicts with the given condition that the period is $16\pi$.

which gives me,
$11$=$a sin 32π$+$c$
$5$=$-a sin$$\left[\dfrac {32π}{3}\right]$+$c$

You're using way too many ## symbols for your LaTeX. All you need are two at the beginning and two more at the end of the expression.

and from the first equation, $sin 32π=0$, and $sin$$\dfrac {32π}{3}$=$\dfrac {\sqrt 3}{2}$

Are you certain that my earlier attempt is not correct?

Yes. I have checked my work.

 

[chwala]

Homework Statement:: See attached.
Relevant Equations:: trigonometry

This is the question...

View attachment 300644
My attempt on part (i),
$b=\dfrac {16π}{2π}=8$
$11=a sin 32π+c$
$c=11$
$5=-a\frac {\sqrt 3}{2} +11$
$10=-a\sqrt 3+22$
$12=a\sqrt 3$
$a=\dfrac {12}{\sqrt 3}$

Is this correct? Thanks...

$b=\dfrac {2π}{16π}=\dfrac {1}{8}$
$11=a+c$
$5=-0.5+c$

$a=4$ and $c=7$

 

[Mark44]

Yes, that's more like it!

 

[SammyS]

$5=-0.5+c$

I think you meant to write: $5=-0.5a+c$

 

[chwala]

I think you meant to write: $5=-0.5a+c$

Absolutely thanks.

 

[chwala]

I think it is

Lol...did you go through the entire thread?

 

[Mark44]

Lol...did you go through the entire thread?

Apparently not...

 

[neilparker62]

$b=\dfrac {2π}{16π}=\dfrac {1}{8}$
$11=a+c$
$5=-0.5+c$

$a=4$ and $c=7$

https://www.desmos.com/calculator/irvvv0gm7b