# $\cos x=12/13$, 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

• $\cos^2 x+\sin^2 x=1$. – André Nicolas Feb 26 '16 at 21:19

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 $\sin x = \frac{y}{13}$.

• 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? – jacobimmugatu Feb 26 '16 at 21:44
• @jacobimmugatu: You forgot to take the square root of $25$. – Brian Tung Feb 26 '16 at 21:47

$\cos x =\frac{12}{13}$ 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

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

Use the formula $\sin(2x)=2\sin x\cos x=2\sqrt{1-\cos^2x}\cos x$

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