How do you use the sum and difference identity to evaluate sin((7pi)/12)?

Aug 20, 2016

$\frac{\sqrt{2 + \sqrt{3}}}{2}$

Explanation:

Trig unit circle -->
$\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{12} + \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{12}\right)$
Find $\cos \left(\frac{\pi}{12}\right)$ by using the trig identity:
$\cos 2 a = 2 {\cos}^{2} a - 1$
$\cos \left(\frac{2 \pi}{12}\right) = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} = 2 {\cos}^{2} \left(\frac{\pi}{12}\right) - 1$
$2 {\cos}^{2} \left(\frac{\pi}{12}\right) = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$
${\cos}^{2} \left(\frac{\pi}{12}\right) = \frac{2 + \sqrt{3}}{4}$
$\cos \left(\frac{\pi}{12}\right) = \pm \frac{\sqrt{2 + \sqrt{3}}}{2.}$
Since cos (pi/12) is positive, take the positive answer.
Finally,
$\sin \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$

Aug 20, 2016

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

Explanation:

We have, $\sin \left(\pi - \theta\right) = \sin \theta$.

$\Rightarrow \sin \left(7 \frac{\pi}{12}\right) = \sin \left(\pi - 5 \frac{\pi}{12}\right) = \sin 5 \frac{\pi}{12}$.

Now, $\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$.

Taking, $A = \frac{\pi}{4} , \mathmr{and} , B = \frac{\pi}{6}$,

we have, $A + B = \frac{\pi}{4} + \frac{\pi}{6} = 3 \frac{\pi}{12} + 2 \frac{\pi}{12} = 5 \frac{\pi}{12}$.

$\therefore \sin 5 \frac{\pi}{12} = \sin \left(\frac{\pi}{4} + \frac{\pi}{6}\right)$.

$= \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) + \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right)$.

$= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2 \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4}$.

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

Just to match this Answer with that furnished by Respected Nghi N. , we see that,

(sqrt3+1)/(2sqrt2)=sqrt{((sqrt3+1)/(2sqrt2))^2}=sqrt{((3+1+2sqrt3)/8)

$= \sqrt{\frac{4 + 2 \sqrt{3}}{8}} = \sqrt{\frac{2 \left(2 + \sqrt{3}\right)}{8}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$

Enjoy Maths.!