How do you find the exact value of #sin((7pi)/12)#?

Hopefully this helps!

To find the exact value of (\sin\left(\frac{7\pi}{12}\right)), you can use the sum or difference identity for sine. The angle (\frac{7\pi}{12}) can be expressed as the sum of two angles whose sine and cosine values are known exactly, such as (\frac{\pi}{3}) and (\frac{\pi}{4}) because (\frac{7\pi}{12} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}).

The sum identity for sine is: [ \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) ]

Using (\frac{\pi}{4}) for (a) and (\frac{\pi}{3}) for (b), we get: [ \sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{3}\right) + \cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{3}\right) ]

Knowing that (\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}), (\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}), (\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}), and (\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}), we can substitute these values into our equation:

[ = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} ] [ = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} ] [ = \frac{\sqrt{2} + \sqrt{6}}{4} ]

Therefore, (\sin\left(\frac{7\pi}{12}\right) = \frac{\sqrt{2} + \sqrt{6}}{4}).