What is the exact value of sin(pi/12)
Explain how you get it
Explain how one gets what? There is no question in your post.
Are you maybe referring to the contents of your subject line...? If so, then try using the same sort of technique which was provided to you here.
If you get stuck, please reply showing how far you have gotten, starting with how you turned 1/12 into a sum or difference of 1/2, 1/3, 1/4, and / or 1/6, and showing which trig identity you are using. Thank you!
Hello,
U have this formula:
$\displaystyle sin^2(x)=\frac{1-cos(2x)}{2}\Rightarrow sin(x)=\sqrt{\frac{1-cos(2x)}{2}}$
$\displaystyle sin(\frac{\pi}{12})=\sqrt{\frac{1-cos(\frac{\pi}{6})}{2}}$
$\displaystyle sin(\frac{\pi}{12})=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$
Also u can have the next method
$\displaystyle sin(\frac{\pi}{4}-\frac{\pi}{6})=sin(\frac{\pi}{4})cos(\frac{\pi}{6} )-sin(\frac{\pi}{6})cos(\frac{\pi}{4})$
$\displaystyle sin(\frac{\pi}{4}-\frac{\pi}{6})=\frac{\sqrt{2}}{2}*\frac{1}{2}-\frac{\sqrt{3}}{2}*\frac{\sqrt{2}}{2}$
Have a nice day,
Hush_Hush.