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Real solutions to

$$\cot^{-1}(x)=\sin^{-1}(x)$$

I found this problem in an exam years ago and I solved it using geometry. The first mistake I made was assuming $\cot^{-1}(x)=\cfrac{1}{\tan^{-1}(x)}$, which i realized quickly enough. I took a $\theta$ such that $\theta=\sin^{-1}(x)=\cot^{-1}(x)\implies \cot \theta=\sin \theta=x$ . This right angled triangle satisfies the condition.enter image description here

Using Pythagoras' Theorem on the triangle and one has the equation $$x^4+x^2-1=0\implies x=\pm\sqrt{\varphi-1} \ \quad \text{where} \ \varphi \text { is the Golden Ratio, } \ \varphi = \cfrac{1+\sqrt 5}2 $$

I gave this question to my teacher in school after the exam and he solved it using calculus and stuffs. I can't really remember (It was 6 years ago) but he clearly didn't use my method.

I would like if someone can show me (an)other way(s) of solving this question.

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Is there a typo in your solution? As written, $1-\varphi$ is negative, so it shouldn't be under a radical. Maybe you mean the smaller golden ration, $\varphi=\frac{\sqrt{5}-1}{2}$. –  alex.jordan Nov 15 '12 at 22:59
    
@alex.jordan thanks! I fixed it! –  user31280 Nov 16 '12 at 0:25

3 Answers 3

up vote 3 down vote accepted

Given a real $x$ the expression $\cot^{-1}(x)={\rm arccot}(x)$ denotes the unique angle $\alpha\in\ ]0,\pi[\ $ with $\cot\alpha=x$. Similarly, given a real $x\in[{-1},1]$ the expression $\sin^{-1}(x)=\arcsin (x)$ denotes the unique angle $\alpha\in[-{\pi\over2},{\pi\over2}]$ with $\sin\alpha=x$.

Therefore the equation $\cot^{-1}(x)=\sin^{-1}(x)$ talks about a number $x\in[-1,1]$ and implicitly about an angle $\alpha\in\ ]0,{\pi\over2}]$ such that $$x=\cot\alpha=\sin\alpha\ .$$ It follows that $\cos\alpha=\sin^2\alpha=1-\cos^2\alpha$, or $$\cos\alpha={\sqrt{5}-1\over2}\ .$$ From $\sin^2\alpha={\sqrt{5}-1\over2}$ we conclude $x=\sin\alpha=\pm\sqrt{{\sqrt{5}-1\over2}}$; but as $\alpha$ has to lie in the interval $\ ]0,{\pi\over2}]$ the only possible solution is $$x=\sqrt{{\sqrt{5}-1\over2}}\doteq 0.78615\ .$$ Since we arrived at this solution by means of correct but "one-way" algebraic manipulations, and not in possession of a general theory about equations of the given type, we should test the found $x$ whether it actually solves the original equation. I leave this verification to the OP.

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You could apply $\sin$ to both sides: $$\sin(\cot^{-1}x)=x$$ Now it helps to know how to simplify the left side. If the cotangent of an angle is $x$, and that angle is part of a right triangle, then we can scale the triangle so that $x$ is the adjacent length, $1$ is opposite, and thus $\sqrt{1+x^2}$ is the hypotenuse. Taking $\sin$: $$\frac{1}{\sqrt{1+x^2}}=x$$

Squaring and clearing denominators is another way to reduce to your equation of $x^4+x^2=1$

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One could prove it with trigonometric identities (although this is in the same spirit as your geometric route).
Take sines of both sides, to get $\sin(\cot^{-1}(x))=x$. Square both sides and take reciprocals, so: $$\frac{1}{x^{2}}=\frac{1}{\sin^{2}(\cot^{-1}(x))}=\csc^{2}(\cot^{-1}(x))=1+\cot^{2}(\cot^{-1}(x))=1+x^{2}$$ In other words, $$x^{4}+x^{2}-1=0 \implies x^{2}=\varphi-1$$

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