Question

1)sin²2x-cos(π/3-2x)•sin(2x-π/6)=1/4
2)1+2cos2x-4cos(π/6+x)•cos(π/6-x)=0​

Answer 1

Докажите тождества :

1)sin²2x-cos(π/3-2x)•sin(2x-π/6)=1/4

2)1+2cos2x-4cos(π/6+x)•cos(π/6-x)=0​

1.

$\displaystyle \sin ^2 2x - \cos \bigg ( \frac{\pi }{3} -2x \bigg )\cdot \sin \bigg ( 2x-\frac{\pi }{6} \bigg )= \frac{1}{4}$

Вспомним формулу приведения

$\bullet$ cos(90° -α) = sinα

Заметим , что

$\cos \bigg ( \dfrac{\pi }{3} -2x \bigg )=\cos \bigg ( \dfrac{\pi }{2} -\bigg (2x+\dfrac{\pi }{6}\bigg) \bigg )=\boxed{\sin \bigg (2x+\dfrac{\pi }{6} \bigg)}$

⇒

$\displaystyle \sin ^2 2x -\sin \bigg (2x+\dfrac{\pi }{6} \bigg) \cdot \sin \bigg ( 2x-\frac{\pi }{6} \bigg )= \frac{1}{4}$

Воспользуемся формулой :

$\bullet ~ \sin \alpha \cdot \sin \beta = \dfrac{1}{2} \Big(\cos (\alpha -\beta ) - \cos (\alpha +\beta )\Big)$

$\displaystyle \sin \bigg (2x+\dfrac{\pi }{6} \bigg) \cdot \sin \bigg ( 2x-\frac{\pi }{6} \bigg ) =\\\\\\= \frac{1}{2 } \bigg ( \cos \Big ( 2x + \frac{\pi }{6}-2x +\frac{\pi }{6} \Big) - \cos \Big ( 2x + \frac{\pi }{6}+2x -\frac{\pi }{6} \Big) \bigg) = \frac{1}{2 } \cdot ( \cos 60^{\circ}-\cos 4x ) = \\\\\\\ = \boxed{\frac{1}{4} - \frac{\cos 4x }{2}}$

$\displaystyle \sin ^2 2x - \bigg (\frac{1}{4} + \frac{\cos 4x }{2} \bigg )= \frac{1}{4}$

$\bullet ~ \sin^2 a = \dfrac{1- \cos 2a}{2} \Rightarrow \boxed{ \sin ^2 2a = \dfrac{1 - \cos 4a}{2} }$

$\displaystyle \dfrac{1 - \cos 4x}{2}-\bigg (\frac{1}{4} - \frac{\cos 4x }{2} \bigg ) = \frac{1}{4}\\\\\\ \frac{1 - \cos 4x}{2 }+ \frac{\cos 4x}{2} - \frac{1}{4} = \frac{1}{4} \\\\\\ \frac{1}{2}- \frac{1}{4} = \frac{1}{4} \\\\\\\ \frac{1}{4} = \frac{1}{4} ~ \checkmark$

Доказано.

2.

$\displaystyle 1+ 2\cos 2x - 4\cos \bigg ( \frac{\pi }{6}+ x \bigg ) \cos \bigg ( \frac{\pi }{6}- x \bigg ) =0$

Воспользуемся формулой :
$\bullet ~ \cos \alpha \cdot \cos \beta = \dfrac{1}{2} \Big(\cos (\alpha -\beta ) + \cos (\alpha +\beta )\Big)$

$\displaystyle\cos \bigg ( \frac{\pi }{6}+ x \bigg ) \cos \bigg ( \frac{\pi }{6}- x \bigg ) = \frac{1}{2} \bigg ( \cos \Big ( \frac{\pi }{6}+ x - \frac{\pi }{6} + x\Big)+ \cos \Big ( \frac{\pi }{6}+ x + \frac{\pi }{6} - x\Big) \bigg ) = \\\\\\ = \frac{1}{2} \Big ( \cos 60^{\circ}+ \cos 2x \Big) = \frac{1}{4}+ \frac{\cos 2x }{2}$

⇒

$\displaystyle 1+ 2\cos 2x - 4 \bigg ( \frac{1}{4}+ \frac{\cos 2x }{2} \bigg ) =0 \\\\ 1 + 2\cos 2x - 1 - 2\cos 2x = 0 \\\\0 = 0$

Доказано.