Question

упростить выражение
cos^2(a-(pi/6))-cos^2(a+(pi/6))​

Answer 1

Ответ:

$\cos^2\alpha+\frac{1}{2}$

Объяснение:

$\cos\left(\alpha-\frac{\pi}{6}\right)=\cos\alpha\cos\frac{\pi}{6}+\sin\alpha\sin\frac{\pi}{6}=\frac{\sqrt{3}}{2}\cos\alpha+\frac{1}{2}\sin\alpha$

$\cos\left(\alpha+\frac{\pi}{6}\right)=\cos\alpha\cos\frac{\pi}{6}-\sin\alpha\sin\frac{\pi}{6}=\frac{\sqrt{3}}{2}\cos\alpha-\frac{1}{2}\sin\alpha$

$\cos^2\left(\alpha-\frac{\pi}{6}\right)=\frac{3}{4}\cos^2\alpha+\frac{1}{4}\sin^2\alpha+\frac{\sqrt{3}}{4} \cos\alpha\sin\alpha$

$\cos^2\left(\alpha+\frac{\pi}{6}\right)=\frac{3}{4}\cos^2\alpha+\frac{1}{4}\sin^2\alpha-\frac{\sqrt{3}}{4} \cos\alpha\sin\alpha$

$\cos^2\left(\alpha-\frac{\pi}{6}\right)+\cos^2\left(\alpha+\frac{\pi}{6}\right)=\\=\frac{3}{4}\cos^2\alpha+\frac{1}{4}\sin^2\alpha+\frac{\sqrt{3}}{4} \cos\alpha\sin\alpha+\frac{3}{4}\cos^2\alpha+\frac{1}{4}\sin^2\alpha-\frac{\sqrt{3}}{4} \cos\alpha\sin\alpha=\\=\frac{3}{2}\cos^2\alpha+\frac{1}{2}\sin^2\alpha=\cos^2\alpha+\frac{1}{2}\cos^2\alpha+\frac{1}{2}\sin^2\alpha=\cos^2\alpha+\frac{1}{2}(\cos^2\alpha+\sin^2\alpha)=\\=\cos^2\alpha+\frac{1}{2}$