Rearranging

**Alex1313** · June 21st 2010, 06:41 PM

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

**undefined** · June 21st 2010, 06:48 PM

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?

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

or

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

?

**Alex1313** · June 21st 2010, 06:56 PM

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

**undefined** · June 21st 2010, 07:36 PM

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.

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

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

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

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

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

**Alex1313** · June 21st 2010, 07:48 PM

thanks anyways

**undefined** · June 21st 2010, 07:58 PM

Originally Posted by Alex1313

thanks anyways

Oh right, degrees.

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

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

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

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

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

If you really wanted to you could write

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

as

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

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