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

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Answer 1 · 2017-03-30T06:00:00

$t=6/piarccos(T/50-1)$

As $T=50+50cos((pit)/6)$

$cos((pit)/6)=(T-50)/50$

and $(pit)/6=arccos(T/50-1)$

i.e. $t=6/piarccos(T/50-1)$

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