Given $\tan(2a)=2$ and $\frac{3\pi}{2}<a<2\pi$ find value of $\tan(a)$
I first found the values of $\cos(2a)$ and $\sin(2a)$ and then used the half angle formula.
$$\tan(a)=\tan\frac{2a}{2}=\frac{1-\cos(2a)}{\sin(2a)} \implies\frac{5}{2\sqrt 5}\left(1-\frac{\sqrt{5}}5\right)$$
I then simplified that to $\frac{\sqrt{5}}{2}-1$
Then because it is in fourth quadrant I multiplied by negative 1 and got:
$$1-\frac{\sqrt{5}}{2}$$
Am I doing something wrong, or am I correct?
On
Just use the double-angle identity. $$\tan(2a) = \frac{2\tan{a}}{1-\tan^2(a)}$$
$$\implies \frac{2\tan{a}}{1-\tan^2(a)} = 2$$
$$\implies 2\tan{a} = 2-2\tan^2{a} \implies \tan{a} = 1-\tan^2{a} \implies \tan^2{a}+\tan{a}-1 = 0$$
Set $t = \tan{a}$ and solve for $t$.
$$t^2+t-1 = 0$$
$$t = \frac{-1\pm\sqrt{5}}{2}$$
Plug in $\tan{a}$.
$$\tan{a} = \frac{-1\pm\sqrt{5}}{2}$$
But $\frac{3\pi}{2} < a < 2\pi$, so $\tan{a} < 0$.
$$\tan{a} = \frac{-1-\sqrt{5}}{2}$$
On
You made two mistakes, but your method does work.
If $\tan(2a)=2$, then either
Case 1: $0< 2a +2 n \pi<\pi/2$, $\cos(2a)=1/\sqrt{5}$, and $\sin(2a)= 2/\sqrt{5}$ (first quadrant), or
Case 2: $\pi<2a+2 n \pi<3 \pi/2$, $\cos(2a)=-1/\sqrt{5}$, and $\sin(2a)= -2/\sqrt{5}$ (third quadrant)
for some integer $n$.
The problem stated that $3 \pi/2 < a < 2 \pi$, so $3 \pi < 2 a < 4 \pi$ and that puts us in Case 2 with $n=-1$.
Now using your method $$\tan(a)=\frac{1 - \cos(2 a)}{\sin(2 a)} = \frac{1+1/\sqrt{5}}{-2/\sqrt{5}} = - \frac{\sqrt{5}+1}{2}. $$
hint
use the well known formula
$$\tan(2a)=\frac{2\tan(a)}{1-\tan^2(a)}=2$$
and $$\tan(a)<0.$$
You need to solve
$$\tan^2(a)+\tan(a)-1=0$$
to find $$\tan(a)=\frac{-1-\sqrt{5}}{2}$$