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I’m trying to figure out the value for $\sin(x)$ when $\tan(x) = 3$. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm doing wrong?

1.) $\tan(x) = 3$, then $\frac{\sin(x)}{\cos(x)} = 3$.

2.) Then $\cos(x) = \frac{1}{3}\sin(x)$

3.) $\sin^2 (x) + \left(\frac{1}{3}\sin(x)\right)^2 = 1$ //Pythagorean identity substitution.

4.) $\left(\frac{4}{3}\sin(x)\right)^2 = 1$ //Combining like terms

5.) $\frac{16}{9}\sin^2(x) = 1$ //Square the fraction so I can move it later.

6.) $\sin^2(x) = \frac{1}{\frac{16}{9}}$ //Divide both sides by $\frac{16}{9}$

7.) $\sin^2(x) = \frac{1}{1} * \frac{9}{16} = \frac{9}{16}$ //divide out the fractions

8.) $\sin(x) = \pm \frac{3}{4}$ //square root both sides.

So $\sin(x) = \pm \frac{3}{4}$ but the book says this is wrong. Any help is much appreciated.

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    $\begingroup$ The arithmetic in 4 is wrong. $\endgroup$ – TorsionSquid Dec 1 '17 at 23:02
  • $\begingroup$ Freshman's dream! $(a+b)^2 \ne a^2 + b^2$ $\endgroup$ – Doug M Dec 1 '17 at 23:44
  • $\begingroup$ Lol! I wish I were a freshman. I'd be 26 years younger. Its just my math skills that are very young :) Your comment is a great reminder. Thank you! $\endgroup$ – maybedave Dec 2 '17 at 0:25
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If $\tan(x)=3$, then $\tan^2(x)=9$. This means that $\frac{\sin^2x}{1-\sin^2x}=9$. So, $\sin^2(x)=\frac9{10}$; in other words (at least if we're on the first quadrant), $\sin(x)=\frac3{\sqrt{10}}$.

Your error lies in item 4: $\sin^2(x)+\left(\frac1{3\sin(x)}\right)^2=\frac{10}9\sin^2(x)$.

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In general, if $\tan(x) = a$ then $a^2 =\tan^2(x) =\dfrac{\sin^2(x)}{\cos^2(x)} =\dfrac{\sin^2(x)}{1-\sin^2(x)} $ so $\dfrac1{a^2} =\dfrac{1-\sin^2(x)}{\sin^2(x)} =\dfrac{1}{\sin^2(x)}-1 $ so $\dfrac{1}{\sin^2(x)} =\dfrac1{a^2}+1 =\dfrac{1+a^2}{a^2} $ so $\sin^2(x) =\dfrac{a^2}{1+a^2} $ or $\sin(x) =\dfrac{a}{\sqrt{1+a^2}} $.

Similarly, $a^2 =\tan^2(x) =\dfrac{\sin^2(x)}{\cos^2(x)} =\dfrac{1-\cos^2(x)}{\cos^2(x)} =\dfrac{1}{\cos^2(x)}-1 $ so $\dfrac{1}{\cos^2(x)} =1+a^2 $ so $\cos(x) =\dfrac1{\sqrt{1+a^2}} $.

Now, try this for hyperbolic functions.

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Another way to think of it is visualizing SOH-CAH-TOA. (I pronounce it "so-cuh-tow-uh".)

Use the relationships on a particular triangle to get the answer.

Tangent is opposite over adjacent (TOA), so choose your vertical side to be $3$ and your horizontal side to be $1$. This right triangle has the correct value for the tangent of the larger acute angle. (One vertex is on the origin, and the tall right triangle is in the first quadrant.)

This means the hypotenuse is $\sqrt{3^2+1^2} = \sqrt{10}$.

Now, for the sine, you take opposite over hypotenuse (SOH): $3/\sqrt{10}$.

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