Express trigonometric expressions in terms of one trigonometric function What is the general process to solve problems such as this:
I'm preparing for this type of exam problem. 

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 A: (Answered before the edit.) For the first question: If $x=\sin \theta $, then $\cos ^{2}\theta =1-\sin ^{2}\theta =1-x^{2}$ and $\cos \theta =\pm \sqrt{1-x^{2}}$. So $$\sin 2\theta =2\sin \theta \cos \theta =\pm 2x\sqrt{1-x^{2}}.\tag{1}$$
As for the second question divide $2\sin \theta \cos \theta$ by $1=\sin ^{2}\theta +\cos ^{2}\theta$ and express both numerator and denominator in terms of $\tan \theta =x$:
$$\begin{eqnarray*}
\sin 2\theta  &=&2\sin \theta \cos \theta =\frac{2\sin \theta \cos \theta }{\sin ^{2}\theta +\cos ^{2}\theta } \\
&=&\frac{\dfrac{2\sin \theta \cos \theta }{\cos ^{2}\theta }}{\dfrac{\sin
^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta }}=\frac{2\tan \theta }{1+\tan
^{2}\theta } \\
&=&\frac{2x}{1+x^{2}}\tag{2}.
\end{eqnarray*}$$
Added: apply the same technique to $\cos 2\theta=\cos^2 \theta-\sin^2 \theta$ to obtain
$$\cos 2\theta=\frac{1-x^2}{1+x^2}.\tag{3}$$
ADDED 2: To derive $(2)$ and $(3)$ you can draw a right triangle with horizontal side (cathetus) $1$, vertical side (cathetus) $x$,  hypotenuse $\sqrt{1+x^2}$ and angle $\theta$ between the hypotenuse and the horizontal cathetus. 
This proves the following 
Theorem: all the direct trigonometric functions of the double-angle $2\theta$ can be expressed in terms of the tangent of the angle $\theta$. 
As a consequence  all the direct trigonometric functions of the angle $\theta$ can be expressed in terms of the tangent of the half angle $\theta/2$.
A: $$ x=tan\theta $$
$$ x^2+1=tan^2\theta+1 = sec^2\theta $$
$$ sin2\theta = 2sin\theta cos\theta = 2tan\theta cos^2\theta = 2tan\theta sec^{-2}\theta = \frac{2x}{x^2+1} $$
