0
$\begingroup$

Can anyone invert

$y = a \cos(x) + b \sin(2x)$

to give $x = f(y)$? An exchange on 29 June 2017 said this was possible but I cannot find the solution. Also, is

$y = a \sin(x) + b \sin(2x)$

invertible in the same way?

Many thanks.

$\endgroup$
2

1 Answer 1

0
$\begingroup$

Notice that the inverse trigonometric functions are not actually the inverse of trigonometric functions. For example, $\sin x$ doesn't actually have an inverse as it fails the horizontal line test. The inverse only exists when we restrict $\sin x$'s domain to be $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

In your function $f(x)=a\cos x+b\sin 2x$, the domain is $\mathbb{R}$, and it's a periodic function, so it clearly does not have an inverse function. The same goes with $g(x)=a\sin x + b\sin 2x$.

$\endgroup$
3
  • $\begingroup$ Of course. Or one could express the result in terms of functions which are themselves repetitive e.g. y = a sin(x) can be inverted to x = asin(y/a). So y = a cos(x) + b sin(x) is easy to invert and y = a tan(x) + b sec(x) can also be inverted, both expressed in repetitive functions. However, I'm also quite happy to find a solution within the domain -pi/2 < x < pi/2. $\endgroup$
    – Steve B
    Apr 12 at 8:30
  • $\begingroup$ David K asked for a link to the original post. Here it is: math.stackexchange.com/questions/2340812/…. $\endgroup$
    – Steve B
    Apr 12 at 8:32
  • $\begingroup$ David K asked for the link to the original post. Here it is: math.stackexchange.com/questions/2340812/…. $\endgroup$
    – Steve B
    Apr 12 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.