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

Question

Elijah Alexander · Accepted Answer

#sin(pi/3 + pi/4) = (sqrt(3) + 1)/(2sqrt(2))#

We can use #sin(A + B) = sinAcosB + cosAsinB#.

#sin(pi/3 + pi/4) = sin(pi/3)cos(pi/4) + sin(pi/4)cos(pi/3)#

#sin(pi/3 + pi/4) = sqrt(3)/2 1/sqrt(2) + (1/sqrt(2))(1/2)#

#sin(pi/3 + pi/4) = (sqrt(3) + 1)/(2sqrt(2))#

We observe that this has the same value as #sin((7pi)/12)#.

Hopefully this helps!

Elijah Alexander · Answer

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