Question

Prove that $sin^{2}(\alpha\ - \frac{\pi}{4}) + sin^{2}(\alpha\ + \frac{\pi}{12}) - sin^{2}(\alpha\ - \frac{\pi}{12}) = \frac{1}{2}$.

Answer 1

Answer:

$\frac{1}{2}$

Step-by-step explanation:

$Formula\:used$

${sin}^2A-{sin}^2B= sin(A+B).sin(A-B)$

${sin}^2(\alpha-\frac{\pi}{4})}+{sin}^2(\alpha+\frac{\pi}{12}})+{sin}^2(\alpha-\frac{\pi}{4}})$

$={sin}^2(\frac{\pi}{4}}-\alpha})+sin(\alpha+\frac{\pi}{12}+\alpha-\frac{\pi}{12}).sin(\alpha+\frac{\pi}{12}-\alpha+\frac{\pi}{12}).$

$=\frac{1}{2}[1-cos2(\frac{\pi}{4}}-\alpha})]+sin2\alpha. sin(\frac{\pi}{6})$

$=\frac{1}{2}[1-cos(\frac{\pi}{2}}-2\alpha})]+sin2\alpha. \frac{1}{2}$

$=\frac{1}{2}-\frac{1}{2}.sin2\alpha+(\frac{1}{2} )sin2\alpha. \\\\=\frac{1}{2}$