This is from an answer key:

Solve$$d(x) = 70\sin(0.65x) + 150 = 100$$ on the interval [0,20]. $$ 0.65x = \sin^{-1}(-\frac 57) + 2\pi k$$ $$x= \frac {{\sin^{-1}(-\frac 57) + 2\pi k}}{0.65}$$ $$x = \frac {3.94 + 2\pi k}{0.65}$$or $$x = \frac {5.94 + 2\pi k}{0.65}$$ I don't see where these values of $\sin^{-1}(-\frac 57)$ come from. I get -0.796 for that value on a calculator. Thanks

  • $\begingroup$ $70\sin(0.65x) = -50$. Do you see it now? $\endgroup$
    – Paul
    Jan 8, 2023 at 19:08
  • $\begingroup$ Yeah, that's just wrong. $\endgroup$ Jan 8, 2023 at 19:12
  • $\begingroup$ @paul did you read the whole question? $\endgroup$ Jan 8, 2023 at 19:13

1 Answer 1


Remember that the inverse sine function, being a function, has to specify only one of the infinitely many angles which have the given value of sine. It does this by requiring that the angle it gives you back is within a certain range, in particular for arcsine, $-\frac\pi 2 \leq \arcsin(x) \leq \frac \pi 2.$ Because of this, if we want all solutions then we need to do a little more work.

So in addition to your $-0.796$ (or coterminally $5.488$) radian solution, there's another set of solutions corresponding with the other set of coterminal angles on the circle with a sine of $-\frac57.$ This is where the $3.94$ comes from, that's basically the supplementary angle of your calculated value, which would have the same sine. (look at the unit circle and think about the intersections it would have with any horizontal line, representing constant y-value)

As far as the $5.94$ goes, I have no idea where that came from and from what I can tell it is just straight-up wrong, I can't even find a reasonable explanation for the mistake that led to it. So that part is still pretty baffling but there is a reason they give two sets of solutions for the inverse sine, and the first one is correct.

Edit: I think I just realized the source of the error, the book was reporting to two decimal places instead of the three we used so they'd have had $5.49$ for the other solution, so it's just a transpositional error, swapping the tenths and hundredths digits. So I think that mystery is solved.

  • $\begingroup$ that explanation was very helpful. There are errors in the book and when I went back and solved for what value instead of 5.94 would give the correct answer I found it to be 5.488. $\endgroup$
    – Rolomoto
    Jan 8, 2023 at 20:33
  • $\begingroup$ @Rolomoto Right, this is what I meant by coterminal angles: they're angle measures offset by a multiple of $2\pi,$ so they point to the same spot on the unit circle, so they have the same trig function values. (also, think I just realized where the error came from, edited the answer to match) $\endgroup$ Jan 8, 2023 at 20:38

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