Determine $\sin a$, $\cos a$, given $\tan a = 2 - \sqrt 3$

Question

Sorry for this stupid question tho, but I'm really stuck with this one. I tried to work according to the principle of $\tan a = \frac{\sin a}{\cos a}$ but the root factor remains an issue for me.

Would someone be able to briefly desribe the probablem to me?

Answer 1

Hint: Write $s= c(2-\sqrt{3})$ and plug it in $s^2+c^2 =1$

where $s=\sin x $ and $c=\cos x$

Answer 2

First draw a triangle and note that $\tan a =\dfrac{\mbox{ opposite side}}{\mbox{adj side}}$, $\sin a=\dfrac{\mbox{ opposite side}}{\mbox{hypotenuse}},\cos a =\dfrac{\mbox{adj side }}{\mbox{hypotenuse}}$

Answer 3

$$\dfrac{\sin A}{2-\sqrt3}=\dfrac{\cos A}1=\pm\sqrt{\dfrac{\sin^2A+\cos^2A}{(2-\sqrt3)^2+1^2}}$$

Now $(2-\sqrt3)^2+1^2=4(2-\sqrt3)$

and $(\sqrt3-1)^2=2(2-\sqrt3)$

Also $2-\sqrt3=\csc30^{\circ}-\cot30^{\circ}=\dfrac{1-\cos30^{\circ}}{\sin30^{\circ}}=\tan15^{\circ}$