Question

cos 4a, если tg 2a = 8

Answer 1

$tg2a=8\\\\\star \; \; 1+tg^2x=\frac{1}{cos^2x}\; \; \Rightarrow \; \; \; cos^2x=\frac{1}{1+tg^2x}\; \; \star \\\\cos^22a=\frac{1}{1+tg^22a}=\frac{1}{1+8^2}=\frac{1}{65}\\\\cos4a=cos^22a-sin^22a=cos^22a-(\underbrace {1-cos^22a}_{sin^22a})=2\, cos^22a-1=\\\\=2\cdot \frac{1}{65}-1=-\frac{63}{65}$