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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 $1-\frac{x^2}{x}$

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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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  • $\begingroup$ It is as simple as that. OP gets the same answer. The book is incorrect $\endgroup$ – DatBoi Aug 29 '20 at 16:27
  • $\begingroup$ @DatBoi Thank you!! $\endgroup$ – maybedave Aug 29 '20 at 16:37
  • $\begingroup$ @Michael Rozenberg. Given the restriction of theta, cos is always positive...BUT...you make a good point in that this is not true of sin. It can be positive or negative. i didn't include this because it was cumbersome to paste this post in the mathjax format but for the value sin, you are right, my solution is not complete. $\endgroup$ – maybedave Aug 29 '20 at 16:37

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