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.
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.
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!