Question

докажите тождества
а)
$1 + sin ^{4} \alpha - cos ^{4} \alpha =2sin ^{2} \alpha$
б)
$sin^ {2} 6 \alpha cos ^{2}6 \alpha (tg ^{2} 6 \alpha + ctg ^{2} 6 \alpha + 2) = 1$
​

Answer 1

$a)\;1+\sin^4\alpha-\cos^4\alpha=1+(\sin^2\alpha-\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha)=\\=1+(\sin^2\alpha-\cos^2\alpha)\cdot1=1+\sin^2\alpha-\cos^2\alpha=\sin^2\alpha+\sin^2\alpha=2\sin^2\alpha$

$b)\;\sin^26\alpha\cos^26\alpha\left(tg^26\alpha+ctg^26\alpha+2\right)=\\=\sin^26\alpha\cos^26\alpha\left(\frac{\sin^26\alpha}{\cos^26\alpha}+\frac{\cos^26\alpha}{\sin^26\alpha}+2\right)=\\=\sin^26\alpha\cos^26\alpha\left(\frac{\sin^46\alpha+\cos^46\alpha}{\sin^26\alpha\cos^26\alpha}+2\right)=\sin^46\alpha+\cos^46\alpha+2\sin^26\alpha\cos^26\alpha=\\=(\sin^26\alpha+\cos^26\alpha)^2=1^2=1$