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apologies if this is an overly simplistic question to answer:

I have the value of $\cos x = \frac{12}{13}$, and I need to find the value of $\sin(2x)$, where $x$ is between $0$ and $90$ degrees (first quadrant).

I have $\sin(2x)=2\sin x(\frac{12}{13})$ but I am stumped trying to find the value of $\sin x$.

Any help would be greatly appreciated

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    $\begingroup$ $\cos^2 x+\sin^2 x=1$. $\endgroup$ – André Nicolas Feb 26 '16 at 21:19
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Because a picture can help:

enter image description here

The triangle is drawn so that $\cos x = \frac{12}{13}$. Use the Pythagorean theorem to find $y$, and then $\sin x = \frac{y}{13}$.

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  • $\begingroup$ ah.. that is what I started with! I used the pythagorean theorem to determine that the opposite side has a value of 25. Then Sinx=25/13? $\endgroup$ – jacobimmugatu Feb 26 '16 at 21:44
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    $\begingroup$ @jacobimmugatu: You forgot to take the square root of $25$. $\endgroup$ – Brian Tung Feb 26 '16 at 21:47
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$\cos x =\frac{12}{13}$ and $\cos^2 x+\sin^2 x=1$ so

$$1-\sin^2x=\left(\frac{12}{13}\right)^2$$

Now, $\color{blue}{\sin(2x)}=2\sin x \cos x=2\sin x\cdot\frac{12}{13}=2\cdot \sqrt{1-\cos^2x}\cdot \frac{12}{13}=2\cdot\sqrt{1-(12/13)^2}\cdot \frac{12}{13}=\color{blue}{\frac{120}{169}}$

Note that I used:

$\sin(2x)=2\sin x \cos x$

and

$\sin x=\sqrt{1-\cos^2 x}$

$$\color{blue}{\sin x}=\sqrt{1-(12/13)^2}=\color{blue}{\frac{5}{13}}$$

Hope it helps

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  • $\begingroup$ Would sin^2x=1-(12/13)^2 be on the right track? $\endgroup$ – jacobimmugatu Feb 26 '16 at 21:40
  • $\begingroup$ @jacobimmugatu You are on the right track. Keep in mind that $0^\circ < x < 90^\circ \implies \sin x > 0$. $\endgroup$ – N. F. Taussig Feb 26 '16 at 21:43
  • $\begingroup$ You meant $$1 - \sin^2x = \left(\frac{12}{13}\right)^2$$ $\endgroup$ – N. F. Taussig Feb 26 '16 at 21:52
  • $\begingroup$ @N.F.Taussig Yes $\endgroup$ – 3SAT Feb 26 '16 at 21:53
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Use the formula $\sin(2x)=2\sin x\cos x=2\sqrt{1-\cos^2x}\cos x$

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    $\begingroup$ Check your last formula: $\sin(2x) = 2\sin x\cos x \neq 2\sqrt{1 - \cos^2x}\sin x$. $\endgroup$ – N. F. Taussig Feb 26 '16 at 21:28
  • $\begingroup$ @HuanXu: Fixed it for you. $\endgroup$ – Brian Tung Feb 26 '16 at 21:32

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