Question

Find the exact value: sin^-1(sin(6π/7))
Not asking for the answer, just asking for some guidance!

Answer 1

Hey Nat!
These inverse functions are a little weird.
In order for them to be a `function`, we restrict their range.
Inverse sine can only produce an angle in Q1 and the quadrant before Q1.

Answer 2

Normally, when we take the composition of a function and it's inverse,
\(\large\rm f^{-1}(f(x))\)
we expect the function and inverse to sort of "undo" one another,
so we simply get back the argument,\[\large\rm f^{-1}(f(x))=x\]

We would like to think of our trig problem in this way,\[\large\rm f^{-1}\left(f\left(\frac{6\pi}{7}\right)\right)=\frac{6\pi}{7}\]But unfortunately it's not going to work that way because 6pi/7 is in Q4, not the Quadrant that we need.

Answer 3

The problem will still work out largely the same though,
we'll simply have to find the angle which is `co-terminal` with 6pi/7.

\(\large\rm \frac{6\pi}{7}-2\pi=?\)
We need to "unwind" one full rotation to get to the correct quadrant.

Answer 4

@zepdrix 6π/7 - 2π = -8π/7 ... I don't think this is it, right?

Answer 5

@zepdrix also thank you for answering, i definitely appreciate it :)

Answer 6

Ohhh I botched that up, sorry.
I was thinking 6pi/7 is in Q4, but it's not, it's in Q2.
So to find our reference angle we subtract our angle from 180.
\(\large\rm \pi-\frac{6\pi}{7}=\frac{\pi}{7}\)

Answer 7

\[\large\rm \sin\left(\frac{6\pi}{7}\right)\quad=\quad \sin\left(\frac{\pi}{7}\right)\]