<s>[INST] If $\tan(\theta$) = $\frac{x}{7}$ for -$\frac{\pi}{2} < \theta < \frac{\pi}{2}$, find an expression for $\sin(2\theta$) in terms of $x$. I'm trying to solve this and am getting a very different answer from the book.  Can someone please help.
I know that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.  So I have to take $\tan(\theta) = \frac{x}{7}$ and find values for sin and cos in terms of $x$.  So here goes:
$\frac{\sin(\theta)}{\cos(\theta)} = \frac{x}{7}$
Then $\cos(\theta) = \frac{7\sin(\theta)}{x}$ and $\sin(\theta) = \frac{x\cos(\theta)}{7}$
Now to find cos and sin in terms of $x$ I plug in values into $\sin^2(\theta) + \cos^2(\theta) = 1$
$\left(\frac{x\cos(\theta)}{7}\right)^2 + \cos^2(\theta) = 1$ then $\cos^2(\theta) = \frac{49}{x^2+49}$
$\sin^2(\theta) + \left(\frac{7\sin(\theta)}{x}\right)^2 = 1$ then $\sin^2(\theta) = \frac{x^2}{x^2+49}$
Now plug these values into $2\sin(\theta)\cos(\theta)$ and I get $2\left(\frac{x}{\sqrt{x^2+49}}\right)\left(\frac{7}{\sqrt{x^2+49}}\right)$
This is very different than the answer in the book and the graphs are different too.  Can someone please tell me what I'm doing wrong?  The book answer is $1-\frac{x^2}{x}$
 [/INST] Your solution is not full.
For example, you can not say immediately that $\sin\theta=\frac{x}{\sqrt{49+x^2}}$ because $|\sin\theta|=\frac{|x|}{\sqrt{49+x^2}}.$
We can make the following.
$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{\frac{2x}{7}}{1+\frac{x^2}{49}}=\frac{14x}{49+x^2}.$$
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