$\cos x = 12/13$ ,where $0
Given $\cos x = 12/13$, where $0<x<90^\circ$, find the value of $\sin(2x)$.
I can't do this question, can you teach me? 
 A: If $\cos x=\frac{12}{13}$ and $x$ is an acute angle, $\sin x=\frac{5}{13}$, since $(5,12,13)$ is a well-known Pythagorean triple, and
$$ \sin(2x) = 2\sin(x)\cos(x) = \frac{120}{169}$$
by the sine duplication formula.
A: use the two trigonometric identities
$$\sin 2x=2\sin x\cos x$$
$$ \sin x=\sqrt{1-\cos^2x}$$
A: Hints: 1) Can you find $\sin x$?
2) Once you know $\sin x$, do you know the expansion formula for $\sin 2x$ in terms of $\sin x$ and $\cos x$?
A: Draw a right angled triangle. Note that using Pythagoras thm 
the Adjacent side= 12, Opposite side= 5, Hypotenuse =13. 
$$ \sin 2 (..) = 2 \cdot 5 \cdot 12 /13^2 =120/169 $$                
A: Draw a right triangle with adjacent side $12$ and hypotenuse $13$ (see figure 1). This triangle will have an angle ($x$, or alternatively $\angle BAC$) with a cosine of $\frac{12}{13}. $Try to find side $\overline{BC}$ to make the calculation of the other trigonometric ratios easier (especially $\sin x$ <== hint hint). The Pythagorean Theorem works well here.
Next, draw a duplicate triangle underneath it (see figure 1 again) and use the law of sines to show that
$$\frac{\sin 2x}{BD} = \frac{\sin\measuredangle B}{AD}$$.
Hint: Do you see why $\sin \measuredangle B = \cos x$?
Can you generalize the preceding to derive the following identity?
$$\sin 2x = 2\sin x\cos x$$
Figure 1

