Inversion of Trigonometric Equations

Question

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.

Thanks!

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*}