If $T = 50 + 50 cos (\frac{π t}{6})$ $T=50+50cos((pit)/6)$ , express $t$ $t$ as a function of $T$ $T$ ?

Question

1299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-30T06:00:00

$t = \frac{6}{π} arccos (\frac{T}{50} - 1)$ $t=6/piarccos(T/50-1)$

As $T = 50 + 50 cos (\frac{π t}{6})$ $T=50+50cos((pit)/6)$

$cos (\frac{π t}{6}) = \frac{T - 50}{50}$ $cos((pit)/6)=(T-50)/50$

and $\frac{π t}{6} = arccos (\frac{T}{50} - 1)$ $(pit)/6=arccos(T/50-1)$

i.e. $t = \frac{6}{π} arccos (\frac{T}{50} - 1)$ $t=6/piarccos(T/50-1)$

If T=50+50cos(πt6)T=50+50cos((pit)/6), express tt as a function of TT?