# Thread: Application - Rewriting A Trigonometric Function

1. ## Application - Rewriting A Trigonometric Function

Acceleration due to gravity, "g", is considered to be $9.8 \frac {m}{s^2}$. However, changes in latitude alter "g" according to the following model:

$g \approx 9.78049(1+.005288\sin^2{\theta} - .000006\sin^2{2\theta})\frac {m}{s^2}$.

a) Rewrite "g" in terms of power $\sin{\theta}$ only.
I know some of the trigonometric identities but I keep getting stuck on rewriting it in terms or $\sin{\theta}$. Help?

b) Find the latitude of Denver, Colorado and determite "g".
I'll be able to do this once I finish part (a).

(c) Find the percent change between the value calculated in part (b) above and the standard value of 9.8.

I'll be able to do this once I finish part (a) and (b).

2. Originally Posted by chrozer
Acceleration due to gravity, "g", is considered to be $9.8 \frac {m}{s^2}$. However, changes in latitude alter "g" according to the following model:

$g \approx 9.78049(1+.005288\sin^2{\theta} - .000006\sin^2{2\theta})\frac {m}{s^2}$.

a) Rewrite "g" in terms of power $\sin{\theta}$ only.
I know some of the trigonometric identities but I keep getting stuck on rewriting it in terms or $\sin{\theta}$. Help?

b) Find the latitude of Denver, Colorado and determite "g".
I'll be able to do this once I finish part (a).

(c) Find the percent change between the value calculated in part (b) above and the standard value of 9.8.

I'll be able to do this once I finish part (a) and (b).
Change $sin(2\theta) = 2sin(\theta)cos(2\theta)$

and then use $cos^{2}(\theta) = 1 - sin^{2}(\theta)$
Then it will be in terms of $sin(\theta)$

3. Originally Posted by bobby
Change $sin(2\theta) = 2sin(\theta)cos(2\theta)$

and then use $cos^{2}(\theta) = 1 - sin^{2}(\theta)$
Then it will be in terms of $sin(\theta)$
But isn't $\sin^2{\theta}$ not in terms of $sin(\theta)$?

4. Originally Posted by chrozer
But isn't $\sin^2{\theta}$ not in terms of $sin(\theta)$?
$\sin^2{\theta}$ is in terms of $\sin(\theta)$

It's just the $\theta$ bit that matters.

5. Originally Posted by bobby
$\sin^2{\theta}$ is in terms of $\sin(\theta)$

It's just the $\theta$ bit that matters.
Ok I see. Thanks alot.