I've been playing around with finding the domain-restricted inverses of trigonometric equations using the inverse trigonometric equations. One of the easier formulas I came up with was the formula for the inverse $$a\cos^2x+b\sin^2x$$ The process I used to invert this was to use the pythagorean identities to turn it into a single trigonometric function: $$=a(\cos^2x+\sin^2x)+(b-a)\sin^2x$$ $$=a(1)+(b-a)\sin^2x$$ $$=a+(b-a)\sin^2x$$ and so now I can easily say that the partial inverse is $$\arcsin\sqrt{\frac{x-a}{b-a}}$$ The other form that I managed to find an inversion formula for was the form $$a\cos x+b\sin x$$ My strategy for this one was to use the sum-angle identity after manipulating the expression a bit: $$=\sqrt{a^2+b^2}\bigg(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\bigg)$$ $$=\sqrt{a^2+b^2}\bigg(\cos x\sin\arcsin\frac{a}{\sqrt{a^2+b^2}}+\sin x\cos\arccos\frac{b}{\sqrt{a^2+b^2}}\bigg)$$ $$=\sqrt{a^2+b^2}\bigg(\cos x\sin\arcsin\frac{a}{\sqrt{a^2+b^2}}+\sin x\cos\arcsin\frac{a}{\sqrt{a^2+b^2}}\bigg)$$ and now I use the sum-angle formula to reduce this to $$=\sqrt{a^2+b^2}\sin\bigg(x+\arcsin\frac{a}{\sqrt{a^2+b^2}}\bigg)$$ and now we can easily find that the inverse is $$\arcsin\bigg(\frac{x}{\sqrt{a^2+b^2}}\bigg)-\arcsin\bigg(\frac{a}{\sqrt{a^2+b^2}}\bigg)$$ However, the last one I've been working with is giving me a little bit of trouble. I can't figure out how to invert $$a\cos x+b\sin^2 x$$ and based on the shapes of its graphs, I suspect some forms can't even be partially inverted this way.

Am I on a wild goose chase? If not, does anybody have any hints?

One more question - does anybody know of other interesting expressions like my two examples that can be inverted? I really enjoy the puzzle of manipulating these expressions to get an inverse, but I don't want to waste my time on any impossible ones.


  • $\begingroup$ In real arcsin t the must be -1<= t <= 1 a/Sqrt[a^2+b^2] may satisfy or may not depending on a and b $\endgroup$
    – Raffaele
    Jun 29, 2017 at 15:34
  • $\begingroup$ @Raffaele Yes, I know, it doesn't cover all of the same domain that the normal $\arcsin$ does. $\endgroup$ Jun 29, 2017 at 15:39
  • $\begingroup$ Change $\sin^2$ to be in terms of $\cos^2$. Now you have a polynomial in terms of $\cos x$ so apply the quadratic formula. I expect there's a load of faffing about to decide which root to take to get $\cos x$ but you can then apply $\arccos$ $\endgroup$ Jul 9, 2017 at 1:07
  • $\begingroup$ Graphing $a\cos^2x+b\sin^2x$, $~a\cos x+b\sin x$, and $a\cos x+b\sin^2x$ for some values of $a$ and $b$ shows why this is so much harder than the other two $\endgroup$ Jul 12, 2017 at 21:21

1 Answer 1


Invert the equation $y=a \cos x +b \sin^2 x$ (for $x$)?

Substitute $\sin^2 x = 1- \cos^2 x$, multiply by $-4b$ and complete the square. \begin{eqnarray*} \underbrace{4 b^2 \cos^2 x -4 ab \cos x + a^2}_{(2b \cos x-a)^2} =a^2 +4b^2 -4by \\ x = \cos^{-1} \left( \frac{a \pm \sqrt{a^2+4 b^2 -4 b y}}{2b} \right) \end{eqnarray*} So in keeping with the question (where $x$ and $y$ are swapped ) the formula that you seek is \begin{eqnarray*} \color{red}{\cos^{-1} \left( \frac{a \pm \sqrt{a^2+4 b^2 -4 b x}}{2b} \right)}. \end{eqnarray*}

FURTHER EXAMPLES : Some other nice examples of similar functions that can be inverted are ... \begin{eqnarray*} a \cos x +b \sin 2 x \,\,\,\,\, & \longleftrightarrow & \,\,\,\,\, ? \end{eqnarray*} \begin{eqnarray*} a \tan^2 x +b \sec^2 x \,\,\,\,\, & \longleftrightarrow & \,\,\,\,\, ? \end{eqnarray*} \begin{eqnarray*} a \tan^2 x +b \sec x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \end{eqnarray*} \begin{eqnarray*} a \tan x +b \sec x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \tiny{\text{HINT : convert to $ \cos$ and $\sin$ and then use the $cos$ half} angle formula.} \end{eqnarray*} \begin{eqnarray*} a \cos(hx+j) +b \cos (hx+k) \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \end{eqnarray*} And here are the two from the question: \begin{eqnarray*} a \cos x +b \sin x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \end{eqnarray*} \begin{eqnarray*} a \cos^2 x +b \sin^2 x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \end{eqnarray*} And another . \begin{eqnarray*} a \cos x +b \sec x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \end{eqnarray*}

  • $\begingroup$ I will not up vote you because I asked for a hint, not a full answer. However, I will up vote (and possibly award the bounty) if you can suggest some other similar inversion problems. $\endgroup$ Jul 11, 2017 at 21:48
  • $\begingroup$ He has done that now $\endgroup$ Jul 12, 2017 at 21:45
  • $\begingroup$ @DonaldSplutterwit Thank you, those are very good examples! $\endgroup$ Jul 12, 2017 at 22:19
  • $\begingroup$ @Nilknarf You could replace $\tan$ by $\cot$ and $\sec$ by $\operatorname{cosec}$ (but these would be almost the same.) I am still trying to dream up some more ... let me know find any more ... $\ddot \smile$ $\endgroup$ Jul 12, 2017 at 22:52
  • $\begingroup$ @DonaldSplutterwit Okay! And thanks for the problems, I really appreciate it... I've had about five previous answers to this question that did the same thing you did, but when I refused to up vote and asked for problems, they just deleted their answers and didn't some back. Definitely not deserving of $+150$... I'll probably give it to you, though. :D $\endgroup$ Jul 12, 2017 at 22:54

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.