Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$

Question

How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$?

Note that $\sin^{-1}$ is the inverse sine function.

I'm asking for the solution $x$ for this equation. Please work out the solution.

Answer 1

Let $\sin^{-1}x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2$

$\implies y+\sin^{-1}(1-\sin y)=\dfrac\pi2-y$

$\implies\sin^{-1}(1-\sin y)=\dfrac\pi2-2y$

$\implies1-\sin y=\sin\left(\dfrac\pi2-2y\right)=\cos2y=1-2\sin^2y$

$\implies2\sin^2y-\sin y=0$

Observation :

For real $\sin^{-1}x, -1\le x\le1$

and for real $\sin^{-1}(1-x),-1\le1-x\le1\implies2\ge x\ge0$

$\implies0\le x\le1\implies0\le y\le\dfrac\pi2$

Answer 2

$\bf{My\; Solution::}$ Given $$\sin^{-1}(x)+\sin^{-1}(1-x) = \cos^{-1}(x)$$

$$\displaystyle \sin^{-1}(1-x) = \cos^{-1}(x)-\sin^{-1}(x)\Rightarrow (1-x) = \sin \left[\cos^{-1}(x)-\sin^{-1}(x)\right]$$

Using $$\sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}$$

So $$\displaystyle (1-x) = \sin\left[\frac{\pi}{2}-2\sin^{-1}(x)\right]=\cos(2\sin^{-1}(x)) = \cos(\cos^{-1}(1-2x^2))$$

Using $$2\sin^{-1}(x)=\cos^{-1}(1-2x^2)$$

So $$\displaystyle (1-x) = (1-2x^2)\Rightarrow 2x^2-x = 0$$

So $$\displaystyle 2x^2-x = 0\Rightarrow x = 0\;\;,x = \frac{1}{2}$$

Now Put into Original Equation we get $\displaystyle x = \frac{1}{2}$ and

$x=0$ satisfy the Given equation.

Answer 3

An alternative approach would be to take the cosine of both sides of the equation and then use the compound angle formula for $\cos(A+B)$. This leads to the equation $$\sqrt{1-x^2}\sqrt{1-(1-x^2)}-x(1-x)=x$$

This simplifies quote nicely and you get the same answers as have already been found, i.e. $x=0$ and $x=\frac 12$

Answer 4

My answer is similar to @juantheron.

First we know that the equation is valid for $0\le\theta\le\dfrac{\pi}2,$ for some $\theta$ in a right triangle. Now taking the cosine of both sides we have:

$$x=\cos(\sin^{-1}(x)+\sin^{-1}(1-x).$$

Using the angle sum formula for cosine and the Pythagorean Theorem, we get

\begin{align} x&=\cos(\sin^{-1}(x))\cos(\sin^{-1}(1-x))-\sin(\sin^{-1}(x))\sin(\sin^{-1}(1-x))\\ &=\sqrt{1-x^2}\sqrt{1-(1-x)^2}-x(1-x)\\ &=\sqrt{(1-x^2)(2x-x^2)}-x(1-x)\\ 2x-x^2&=\sqrt{(1-x^2)(2x-x^2)}\\ \end{align} Now, noting that $x=0$ is a solution, the remaining solutions come from \begin{align} \sqrt{2x-x^2}&=\sqrt{1-x^2}\\ 2x&=1\\ x&=\dfrac12. \end{align}

Answer 5

My approach for these kinds of questions is to draw triangles, especially when I meet equations with inverse trigonometric functions. (Although it may lose generality sometimes, it is still a good way to work out some of the solutions.)

So why not draw a triangle for this?

Obviously, when we interpret the equation graphically, $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$

holds if it's an equilateral triangle. So we have

$$\cos^{-1} x=\frac \pi 3\implies x=\cos\frac\pi 3=\frac12$$

It is also easy to see when $x$ tends to $0$, both triangles collapse and $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ must hold.