The value of sin(2x) is
![(120)/(169)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ftzz0tgur1guhq1t9tmix7r6q7vx5s9mn2.png)
Step-by-step explanation:
Given that
![tan x =(12)/(5)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8y978bm3e4g84orew4qhfir1d40a0i7xq3.png)
The formula for
is
![\sin (2 x)=2 \sin x \cos x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/g2tu6x5pdpkz0k0q902z7iv16yqh7o273f.png)
Since,
![\tan x=(o p p)/(a d j)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/j37f60vanop4nup69urq3q4rql1bvualgc.png)
Also, it is given that
![tan x =(12)/(5)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/8y978bm3e4g84orew4qhfir1d40a0i7xq3.png)
Thus,
and
![adj=5](https://img.qammunity.org/2021/formulas/mathematics/middle-school/imn3lvhmkv3vjmbo4qk2hcaw4bd1lgyv35.png)
To find the hypotenuse, let us use the pythagoras theorem,
![\begin{aligned}h y p &=\sqrt{12^(2)+5^(2)} \\&=√(144+25) \\&=√(169) \\&=13\end{aligned}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/q7k41p7vp3j2lzet7bavwc1ark6bizsk24.png)
Now, we can find the value of sin x and cos x.
![\sin x=\frac{\text { opp }}{h y p}=(12)/(13)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/a4bxk2j5s2u5oo476gafgyw0ds283q4hjo.png)
![\cos x=(a d j)/(h y p)=(5)/(13)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/rzxyyy9c9rn6e837a4rmhk8nu8kv696q6n.png)
Now, substituting these values in the formula for sin 2x, we get,
![\begin{aligned}\sin (2 x) &=2 \sin x \cos x \\&=2\left((12)/(13)\right)\left((5)/(13)\right) \\&=(120)/(169)\end{aligned}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/63h8ebt2f9w5585bwffr7b838fzfy8mkzn.png)
Thus, the value of sin(2x) is
![(120)/(169)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ftzz0tgur1guhq1t9tmix7r6q7vx5s9mn2.png)