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I am coming up with the wrong answer.

The problem is $\sin[\arctan(4x)]$

Here are the reciprocal functions I am using $\sin\theta= \frac {1}{\csc\theta}$ and $\cot\theta = \frac{1}{\tan\theta}$

Here is the Pythagorean identity I am using $\csc^2 \theta = 1+\cot^2 \theta$

I am trying to rewrite the problem $\sin[\arctan(4x)]$ in terms of the Pythagorean Identity using the reciprocal identities. $$ \frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac{1}{1+\tan^2 \theta}}$$ This seems to be a stretch, I do not think I am expressing the Pythagorean Identity right, and I am not sure how to either. I did start to try to plug in the numbers.

$$ \frac{1}{\csc^2 \theta} = \frac {1}{1 + \frac {1}{(4x^2)}}$$

$$ \frac{1}{\csc \theta} = \sqrt{1+4x} $$

$$\sin\theta = \frac{1}{\sqrt{1+4x}} $$

I could rationalize the denominator and multiply the by its conjugate to get; $$\frac{\sqrt{1-4x}}{\sqrt{(1+4x)(1-4x)}} $$ But, at this point I know I am wrong. The answer is $$ \frac{4x}{\sqrt{16x^2+1}}$$

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    $\begingroup$ Not sure why you're trying to introduce these reciprocal functions. If you drew out a triangle with opposite 4x and adjacent 1, it finishes this really quickly. $\endgroup$
    – David P
    Commented Feb 11, 2022 at 21:29
  • $\begingroup$ Not sure but they are trig identities. I was wondering how to use them in the problem. Also I didn't know the difference between tangent and hypotenuse. $\endgroup$
    – Benp404
    Commented Feb 11, 2022 at 21:48

3 Answers 3

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I am going to bypass examining your work, because the problem permits a very easy shortcut.

If you have a right triangle, whose tangent is $(4x)$, then the legs are (in effect) $(4x)$ and $(1)$.

Therefore, the hypotenuse is $~\displaystyle \sqrt{16x^2 + 1}.$

Therefore, the sine of this angle must be

$$\frac{4x}{\sqrt{16x^2 + 1}}.$$

Addendum
Responding to the comment of Dan:

Need to show that it's still valid when $x < 0$.

I agree, my oversight.

When $x < 0,$ you can let $y = -x \implies y > 0$.

Then, if $\tan(\theta) = 4x$, you have that $\tan(-\theta) = 4y.$

Then, by the analysis at the start of this answer, $\sin(-\theta) = \displaystyle ~\frac{4y}{\sqrt{16y^2 + 1}} = \frac{-4x}{\sqrt{16x^2 + 1}}.$

Therefore, since $\sin(\theta) = -\sin(-\theta)$, you have that

$\sin(\theta) = \displaystyle \frac{4x}{\sqrt{16x^2 + 1}}.$

Edit
In my opinion, an open issue is whether the Addendum is actually needed. In Analytical Geometry, where the domain of trig functions are angles, you have the issue of whether a right triangle can be constructed, some of whose side lengths are negative.

The analysis at the start of my answer was based on physically constructing the analogous right triangle. I don't know how the issue is being taught in Analytical Geometry, so it is better to add the Addendum, erring on the side of caution.

In Real Analysis (AKA Calculus) the issue is somewhat convoluted, because the domain of the trig functions are real numbers, rather than angles. However, you can interpret the domain of the trig functions to be various arc lengths, with respect to the unit circle.

Under this interpretation, it seems to me that the Addendum is not needed, because you can construct a right triangle [whose hypotenuse is $(1)$] that lies in the $4$th quadrant just as appropriately as constructing a right triangle that lies in the $1$st quadrant. Then, you have constructed a right triangle, one of whose legs is a negative number.

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    $\begingroup$ Thanks, that is much easier. I wasn't sure how to express the hypotenuse. $\endgroup$
    – Benp404
    Commented Feb 11, 2022 at 21:39
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    $\begingroup$ Nitpick: Needs to show that it's still valid when $x < 0$. $\endgroup$
    – Dan
    Commented Feb 11, 2022 at 21:40
  • $\begingroup$ $ hypotenuse^2= 4x^2+1$ I was thinking the tangent was but couldn't be the hypotenuse. $\endgroup$
    – Benp404
    Commented Feb 11, 2022 at 21:46
  • $\begingroup$ @Dan Nice catch, thanks. Addendum added to the end of my answer. $\endgroup$ Commented Feb 11, 2022 at 21:49
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    $\begingroup$ @Benp404 See also the addendum added to the end of my answer. The idea is that when $(4x) < 0$, then you can't really construct the pertinent right triangle, because you can't construct a triangle with side lengths less than zero. $\endgroup$ Commented Feb 11, 2022 at 21:53
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Let's try your idea, but in a slightly different way: it's way simpler to compute $\sin(\arctan(x))$ and then substituting $4x$ for $x$.

The limitation now is $-1\le x\le 1$, so $-\pi/4\le\arctan(x)\le\pi/4$. Set $y=\arctan(x)$. We have $$ \sin^2y+\cos^2y=1 $$ and therefore $$ \sin^2y=\frac{\sin^2y}{\sin^2y+\cos^2y}=\frac{\tan^2y}{\tan^2y+1} $$ But $\tan y=x$ by definition, so $$ \sin^2(\arctan(x))=\frac{x^2}{x^2+1} $$ Thus we have $$ \sin(\arctan(x))=\frac{x}{\sqrt{x^2+1}} $$ by comparing the cases when $x>0$ and $x<0$. Now you can see that the limitation $-1\le x\le 1$ is redundant and the statement holds for every $x$.

With $4x$ in place of $x$: $$ \sin(\arctan(4x))=\frac{4x}{\sqrt{16x^2+1}} $$

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  • $\begingroup$ I was wondering how you go from $sin^2y$. To $sin^2y = \frac{sin^2y}{sin^2y+cos^2y}$ $\endgroup$
    – Benp404
    Commented Feb 12, 2022 at 3:43
  • $\begingroup$ @Benp404 $a=a/1$; it’s the usual trick to express the (squared) sine or cosine in terms of the tangent. $\endgroup$
    – egreg
    Commented Feb 12, 2022 at 8:55
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As egreg said, the condition is redundant. On the other words, for any real $x$,$$\sin(\arctan(4x))=\frac{4x}{\sqrt{16x^2+1}}.$$ I like to prove it by 2 cases. Let $y=\arctan 4x$.

A. When $x\geq 0,$ $$\sin ^{2} y=\tan ^{2} y \cos ^{2} y=16 x^{2}\left(1-\sin ^{2} y\right) \Rightarrow \sin ^{2} y=\frac{16 x^{2}}{16 x^{2}+1} \Rightarrow \sin y=\frac{4 x}{\sqrt{16 x^{2}+1}} $$

Hence $$ \sin (\arctan (4 x))=\frac{4 x}{\sqrt{16 x^{2}+1}}. $$

B. When $x< 0,$ let $z=-x>0,$ then using the result in case A yields $$ \sin (\arctan 4 z)=\frac{4 z}{\sqrt{16 z^{2}+1}}. $$

Now putting $z=-x$ yields $$ \sin (\arctan 4(-x))=\frac{-4 x}{\sqrt{16 (-x)^{2}+1}}. $$ Using the facts that $\sin x$ and $\arctan x $are odd functions, we have$$ -\sin (\arctan (4x))=-\frac{4 x}{\sqrt{16 x^{2}+1}}. $$

$$ \sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}. $$

Therefore we can conclude that for any real $x$, $$ \boxed{\sin (\arctan (4x))=\frac{4 x}{\sqrt{16 x^{2}+1}}}. $$ In general, for any real $x$, $$\boxed{\sin (\arctan (x))=\frac{x}{\sqrt{x^{2}+1}}}$$

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