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What is the general process to solve problems such as this:

I'm preparing for this type of exam problem. enter image description here

For Reference, similar/more advanced problem (with solution):

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Find as many practices as you can and do them different ways to improve speed and accuracy and then try to solve them using different substitutions. Practice, Practice, Practice is about all that can be said. – Amzoti Mar 18 '13 at 0:38
@Amzoti Thanks, much easier than I thought. I put my work above. – Patrick Mar 18 '13 at 0:55
Excellent! Now, try to do it different ways and with different substitutions and then pick up your speed and accuracy and try as many as you can get your hand on! – Amzoti Mar 18 '13 at 0:57
@Amzoti OK, not sure what other substitutions to use. This seems like the quickest way to the answer. – Patrick Mar 18 '13 at 1:00
up vote 2 down vote accepted

(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 $\theta/2$.

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Tavares: Many thanks for thorough explanation. For the 2nd problem, I believe my interpretation is a bit simpler: (and also drew the simple triangle). – Patrick Mar 18 '13 at 3:43
@PatrickNommensen You are welcome! Yes, it is indeed. – Américo Tavares Mar 18 '13 at 13:03

$$ x=tan\theta $$ $$ x^2+1=tan^2\theta+1 = sec^2\theta $$ $$ sin2\theta = 2sin\theta cos\theta = 2tan\theta cos^2\theta = 2tan\theta sec^{-2}\theta = \frac{2x}{x^2+1} $$

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Thanks, but nothing new here. – Patrick Mar 18 '13 at 0:59

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