Question

Помогите доказать тождество, пожалуйста))

Answer 1

Для доказательства использована формула разности квадратов

$a^2-b^2=(a-b)(a+b)$

и тригонометрические тождества:

$\sin^2\alpha +\cos^2\alpha =1\ \ \ \ \ \Leftrightarrow\ \ \ \ \ \cos^2\alpha =1-\sin^2\alpha\\\\tg\;\alpha =\dfrac{\sin\alpha }{\cos\alpha }$

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$1. \\\cos^2\alpha -\sin^2\alpha =1-2\sin^2\alpha \\1-\sin^2\alpha -\sin^2\alpha =1-2\sin^2\alpha \\\boldsymbol{1-2\sin^2\alpha=1-2\sin^2\alpha}\ \ \ \ ~~~\blacksquare$

$2.\\\cos^4\alpha +\sin^2\alpha \cos^2\alpha +\sin^2\alpha =1\\\cos^2\alpha \big(\underbrace{\cos^2\alpha +\sin^2\alpha }_{=1}\big)+\sin^2\alpha =1\\\cos^2\alpha +\sin^2\alpha =1\\~~~~~~~~~~~~~~~~1=1~~~~~~~~~~\blacksquare$

$3.\\\dfrac1{\sin\alpha -\cos\alpha }=\dfrac{\sin\alpha +\cos\alpha }{\sin^4\alpha -\cos^4\alpha }\\\\\dfrac1{\sin\alpha -\cos\alpha }=\dfrac{\sin\alpha +\cos\alpha }{(\sin^2\alpha -\cos^2\alpha)(\underbrace{\sin^2\alpha +\cos^2\alpha}_{=1}) }\\\\\dfrac1{\sin\alpha -\cos\alpha }=\dfrac{\sin\alpha +\cos\alpha }{\sin^2\alpha -\cos^2\alpha }\\\\\dfrac1{\sin\alpha -\cos\alpha }=\dfrac{\sin\alpha +\cos\alpha }{(\sin\alpha -\cos\alpha )(\sin\alpha +\cos\alpha )}$

$\boldsymbol{\dfrac1{\sin\alpha -\cos\alpha }=\dfrac1{(\sin\alpha -\cos\alpha )}}\ \ \ \ \blacksquare$

$4.\\\dfrac{~~~\sin\alpha +tg\;\alpha }{1+\cos\alpha }=tg\;\alpha \\\\\dfrac{\sin\alpha +\frac{\sin\alpha }{\cos\alpha} }{1+\cos\alpha }=tg\;\alpha \\\\\dfrac{\sin\alpha \cdot \cos\alpha +\sin\alpha }{\cos\alpha\cdot(1+\cos\alpha) }=tg\;\alpha \\\\\dfrac{\sin\alpha \cdot (\cos\alpha +1)}{\cos\alpha\cdot(1+\cos\alpha) }=tg\;\alpha \\\\\dfrac{\sin\alpha}{\cos\alpha }=tg\;\alpha \\\\\boldsymbol{tg\;\alpha=tg\;\alpha}\ \ \ \ \ \ \blacksquare$

$5.\\\dfrac{~~~\cos^2\alpha -\sin^2\alpha}{ctg\;\alpha-tg\;\alpha}=\sin\alpha\cos\alpha\\\\\dfrac{\cos^2\alpha-\sin^2\alpha}{\frac{\cos\alpha}{\sin\alpha}-\frac{\sin\alpha}{\cos\alpha}}=\sin\alpha\cos\alpha\\\\\big(\cos^2\alpha-\sin^2\alpha\big):\dfrac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}}=\sin\alpha\cos\alpha\\\\\dfrac{\big(\cos^2\alpha-\sin^2\alpha\big)\cdot\sin\alpha\cos\alpha}{\big(\cos^2\alpha-\sin^2\alpha\big)}}=\sin\alpha\cos\alpha$

$\boldsymbol{\sin\alpha \cos\alpha=\sin\alpha \cos\alpha}\ \ \ \ \ \blacksquare$