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

  • 3
    $\begingroup$ $\cos^2 x+\sin^2 x=1$. $\endgroup$ – André Nicolas Feb 26 '16 at 21:19

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @jacobimmugatu: You forgot to take the square root of $25$. $\endgroup$ – Brian Tung Feb 26 '16 at 21:47

$\cos x =\frac{12}{13}$ and $\cos^2 x+\sin^2 x=1$ so


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:

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


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

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

Hope it helps

  • $\begingroup$ 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

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

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.