# Solving a trig equation with inverse?

How would I solve the following trig equation?

$\arctan(x)+\arcsin(x)=\frac \pi 2$

I am kind of confused on how to solve it.

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We can obviously rearrange it to $$\arcsin(x)=\frac{\pi}{2}-\arctan(x),$$, which is equivalent to $$x=\sin(\frac{\pi}{2}-\arctan(x)).$$ Using the difference-angle formula, we have $$x=\cos(\arctan(x)).$$ Now, viewing $x=\frac{x}{1}$, we can think of $\cos(\arctan(x))$ as the sides of a triangle, in particular, we have $$x=\frac{1}{\sqrt{x^2+1}}.$$ Squaring and multiplying both sides by $x^2+1$ quickly yields the equation $$x^4+x^2-1=0.$$ Set $w=x^2$ so that the equation above becomes $$w^2+w-1=0.$$ The quadratic formula gives us $$x^2=w=\frac{-1\pm\sqrt{1-4(1)(-1)}}{2(1)}.$$ Simplifying the expression and observing that $x^2\geq0$ for all real numbers $x$, we have $$x=\sqrt{\frac{-1+\sqrt{5}}{2}}.$$ I hope this helps!
The difference angle formula for $\sin(\alpha-\beta)$ is $\sin(\alpha)\cos(\beta)-\sin(\beta)\cos(\beta)$. See if you can verify the work from there. –  Clayton Dec 16 '12 at 19:03
With $\arctan(x)$, the legs of the right triangle will be $x$ and $1$ where $x$ is the side opposite your angle, and the leg of length $1$ is adjacent to the angle. Using the Pythagorean Thm, we have the hypotenuse is $\sqrt{x^2+1}$. Checking that cosine is the adjacent leg over the hypotenuse, we have $$\cos\big(\arctan(x)\big)=\frac{1}{\sqrt{x^2+1}}.$$ Does this help at all? –  Clayton Dec 16 '12 at 19:34
$\sin(\frac{\pi}{2}-\arctan(x))=\sin(\frac{\pi}{2})\cos(\arctan(x))-\sin(\arctan‌​(x))\cos(\frac{\pi}{2})$. As you noted above, $\sin(\frac{\pi}{2})=1$ and $\cos(\frac{\pi}{2})=0$, so the second term drops out, while the first one reduces to $\cos(\arctan(x))$. –  Clayton Dec 16 '12 at 19:48
$$x^2=(x)^2=\left(\frac{1}{\sqrt{x^2+1}}\right)^2=\frac{1}{x^2+1}$$ Now see if you can multiply and get what I have above. –  Clayton Dec 16 '12 at 20:08