Addition formula for sine? (sin(arctanx+arctan2x)=root2/2)

[messa]

Hello, I'm working on this problem and I feel like I'm on the right track, but I've gotten stuck. Here is the problem:

Find all solutions to
sin(arctanx+arctan2x)=root2/2

This is the work I've done:

sin(u+v)=(sinu)(cosv)+(cosu)(sinv)
I've constructed two triangles to represent u and v

u: hypotenuse=root 2, opposite =1, adjacent=1
V: hypotenuse=root5, opposite=2, adjacent=1
(1/root2)(1/root5)+(1/root2)(2/root5)

I'm not sure where to take it from here especially since it says find all answers, and what does the root2/2 in the initial question have to do with anything?

 

[soroban]

Re: Addition formula for sine?

Hello, messa!

An unusual problem . . .

Find all solutions to: \(\displaystyle \:\sin(\arctan x\,+\,\arctan2x)\:=\:\frac{\sqrt{2}}{2}\)

Take the inverse sine of both sides:
. . \(\displaystyle \arctan x\,+\,\arctan2x \:=\:\arcsin\left(\frac{\sqrt{2}}{2}\right) \:=\:\left\{\begin{array}{ccc}\frac{\pi}{4}\,+\,2\pi n \\ \frac{3\pi}{4}\,+\,2\pi n\end{array}\)

Take the tangent of both sides:
. . \(\displaystyle \tan(\arctan x\,+\,\arctan2x) \;=\;\left\{\begin{array}{ccc}\tan\left(\frac{\pi}{4}\,+\,2\pi n\right) \\ \tan\left(\frac{3\pi}{4}\,+\,2\pi n\right) \end{array}\)

We have: \(\displaystyle \L\:\frac{\tan(\arctan x)\,+\,\tan(\arctan2x)}{1\,-\,\tan(\arctan x)\cdot\tan(\arctan2x)} \;=\;\pm1\;\;\Rightarrow\;\;\frac{x\,+\,2x}{1\,-\,(x)(2x)} \;=\;\pm1\)


There are two equations to solve:


\(\displaystyle \L(1)\;\frac{3x}{1\,-\,2x^2}\:=\:1\;\;\Rightarrow\;\;2x^2\,+\,3x\,-\,1\:=\:0\)

. . . \(\displaystyle \L x\:=\:\frac{-3\,\pm\sqrt{3^2\,-\,4(2)(-1)}}{2(2)} \:=\:\fbox{\frac{-3\,\pm\,\sqrt{17}}{4}}\)


\(\displaystyle \L(2)\;\frac{3x}{1\,-\,2x^2}\:=\:-1\;\;\Rightarrow\;\;2x^2\,-\,3x\,-\,1\:=\:0\)

. . \(\displaystyle \L x\:=\:\frac{3\,\pm\,\sqrt{(-3)^2\,-\,4(2)(-1)}}{2(2)} \:=\:\fbox{\frac{3\,\pm\,\sqrt{17}}{4}}\)


I'll let you check for extraneous roots . . .

 

[messa]

So, I wasn't on the right track at all?!