How do you find the exact value of sin67.5 degrees?

Jul 2, 2015

We could use the half-angle identity:
$\sin \left(\frac{1}{2} x\right) = \pm \sqrt{\frac{1 - \cos x}{2}}$

Explanation:

${67.5}^{2} = \frac{1}{2} \left({135}^{\circ}\right)$ (Multiply $67.5 \times 2$.)

The formula doesn't tell us whether $\sin {67.5}^{\circ}$ is positive or negative, but, since it is an acute angle we know that the sine is positive. (Be careful of the difference between "sign" and "sine").

We also need $\cos {135}^{\circ}$. (That is the special angle that is ${45}^{\circ}$ in Quadrant II.)

$\cos {135}^{\circ} = - \frac{\sqrt{2}}{2}$.

Now use the formula ans simplify:

$\sin {67.5}^{\circ} = \sin \left(\frac{1}{2} \left({135}^{\circ}\right)\right) = \sqrt{\frac{1 - \cos {135}^{\circ}}{2}}$

$= \sqrt{\frac{1 - \left(- \frac{\sqrt{2}}{2}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$

$= \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

$\sin {67.5}^{\circ} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

Jul 3, 2015

Find sin 67.5

Explanation:

Call sin 67.5 = sin x

$\cos 2 x = \cos 135 = - \frac{\sqrt{2}}{2} = 1 - 2 {\sin}^{2} x$

$2 {\sin}^{2} x = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}$

${\sin}^{2} x = \frac{2 + \sqrt{2}}{4}$

$\sin x = \sin 67.5 = \frac{\sqrt{2 + \sqrt{2}}}{2}$