How would I go on about solving: $\,\quad\cos\Big(\,2\cdot \arctan\big(\frac{1}{2}\big)\Big)$ ?

My attempt was to find $\,v\,$ in $\,\frac{\sin\left(v\right)}{\cos\left(v\right)} = \frac{1}{2}\,$ and substitute that with $\,\arctan\left(\frac{1}{2}\right),\,$ but I end up with $\,\sin\left(v\right) = \sqrt{\frac{1}{5}}\,$ which I cant solve.

Is there any other way to approach the problem?

  • 1
    $\begingroup$ Solving implies that there's an equation, but what you have there is just an expression. Do you mean you just want to simplify it? $\endgroup$ – Brenton Sep 15 '15 at 21:50

So you have $\sin(v) = \sqrt{\frac{1}{5}}$, then $\cos^2(v) = \frac{4}{5}$. Now you just need to calculate $\cos(2v) = 2\cos^2(v) - 1 = \frac{3}{5}$


HINT: Recall the tangent half-angle formula

$$ \cos 2\alpha = \dfrac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} $$


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