$\cos x=12/13$, where $0 \lt x \lt 90$ degrees. What is the value of $\sin(2x)?$
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
$\endgroup$ 13 Answers
$\begingroup$Because a picture can help:
The triangle is drawn so that $\cos x = \frac{12}{13}$. Use the Pythagorean theorem to find $y$, and then $\sin x = \frac{y}{13}$.
$\endgroup$ 2 $\begingroup$$\cos x =\frac{12}{13}$ and $\cos^2 x+\sin^2 x=1$ so
$$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
$\endgroup$ 4 $\begingroup$Use the formula $\sin(2x)=2\sin x\cos x=2\sqrt{1-\cos^2x}\cos x$
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