1. ## Rearranging

Hi, have to rearrange this equation for t, but keep getting it wrong. Can anyone help? Thanks and my gratitude

21=10 sin 10(t-9) + 11

2. Originally Posted by Alex1313
Hi, have to rearrange this equation for t, but keep getting it wrong. Can anyone help? Thanks and my gratitude

21=10 sin 10(t-9) + 11
Hi and welcome!

Which do you mean?

$\displaystyle 21=10 \sin (10(t-9)) + 11$

or

$\displaystyle 21=10 (\sin 10)(t-9) + 11$

?

3. first one. It should be 18. I just want to know how to solve for t with a trig function

4. Originally Posted by Alex1313
first one. It should be 18. I just want to know how to solve for t with a trig function
I don't know why it would be 18.

$\displaystyle 21=10 \sin (10(t-9)) + 11$

$\displaystyle 10=10 \sin (10(t-9))$

$\displaystyle 1= \sin (10(t-9))$

$\displaystyle 10(t-9) = \frac{\pi}{2}+k\cdot2\pi, k\in\mathbb{Z}$

$\displaystyle t=9+\frac{\pi}{20}+k\cdot\frac{\pi}{5}, k\in\mathbb{Z}$

5. thanks anyways

6. Originally Posted by Alex1313
thanks anyways
Oh right, degrees.

$\displaystyle 21=10 \sin (10(t-9)^\circ) + 11$

$\displaystyle 10=10 \sin (10(t-9)^\circ)$

$\displaystyle 1= \sin (10(t-9)^\circ)$

$\displaystyle 10(t-9)^\circ = 90^\circ+k\cdot360^\circ, k\in\mathbb{Z}$

$\displaystyle t=18+k\cdot36, k\in\mathbb{Z}$

If you really wanted to you could write

$\displaystyle 10(t-9)^\circ = 90^\circ+k\cdot360^\circ, k\in\mathbb{Z}$

as

$\displaystyle 10(t-9)^\circ = \sin^{-1}(1)+k\cdot360^\circ, k\in\mathbb{Z}$