Question

С: ) sin π/8×cos π/8= sin²α×cos²α= sin² π/8-cos² π/8= sin π/12×cos π/12= 2tg15°/1-tg²15°= (1-sin )×(1+sin)= sin20°/cos10°= 1+cos8α=

Answer 1

$Sin \frac{ \pi }{8}Cos \frac{ \pi }{8} = \frac{1}{2} *2Sin \frac{ \pi }{8}Cos \frac{ \pi }{8}= \frac{1}{2}Sin \frac{ \pi }{4} = \frac{1}{2}* \frac{ \sqrt{2} }{2}= \frac{ \sqrt{2} }{4}$

$Sin ^{2} \alpha Cos \alpha ^{2} = \frac{1}{4} *4Sin ^{2} \alpha Cos ^{2} \alpha = \frac{1}{4} Sin ^{2}2 \alpha$

$Sin ^{2} \frac{ \pi }{8} -Cos ^{2} \frac{ \pi }{8} =-(Cos ^{2} \frac{ \pi }{8} -Sin ^{2} \frac{ \pi }{8} )=-Cos \frac{ \pi }{4}=- \frac{ \sqrt{2} }{2}$

$Sin \frac{ \pi }{12}Cos \frac{ \pi }{2} = \frac{1}{2}*2Sin \frac{ \pi }{12}Cos \frac{ \pi }{12} = \frac{1}{2} Sin \frac{ \pi }{6}= \frac{1}{2}* \frac{1}{2} = \frac{1}{4}$

$\frac{2tg15 ^{o} }{1-tg ^{2}15 ^{o} } =tg30 ^{o}= \frac{ \sqrt{3} }{3}$

(1 - Sinα)(1 + Sin) = 1 - Sin²α = Cos²α

$\frac{Sin20 ^{o} }{Cos10 ^{o} }= \frac{2Sin10 ^{o} Cos10 ^{o} }{Cos10 ^{o} } =2Sin10 ^{o}$

$1+Cos8 \alpha =2Cos ^{2}4 \alpha$