Question

ТРИГОНОМЕТРИЯ Помогите с этим, пожалуйста, хотя бы с 5-6. Очень нужно, спасибо

Answer 1

5
cosacosb-sinasinb=cos(a+b)
cosπ/5cos2π/15-sinπ/5sin2π/15=cos(π/5+2π/15)=cos(3π/15+2π/15)=сos5π/15=
=сosπ/3=1/2
6
sinacosb-sinbcosa=sin(a-b)
sin5π/24cos3π/8-cos5π/24sin3π/8=sin(5π/24-3π/8)=sin(-π/6)=-1/2

Answer 2

$\frac{tg13+tg32}{1-tg13*tg32}=tg(13+32)=tg45=1\\\\ \frac{tg65-tg35}{1+tg65*tg35} =tg(65-35)=tg30= \frac{ \sqrt{3} }{3} \\\\ \frac{tg111+tg24}{1-tg111*tg24} =tg(111+24)=tg135=tg(90+45)=-Ctg45=-1$$Cos \frac{ \pi }{5} Cos \frac{2 \pi }{15} -Sin \frac{ \pi }{5} Sin \frac{2 \pi }{15}=Cos( \frac{ \pi }{5}+ \frac{2 \pi }{15})=Cos( \frac{3 \pi +2 \pi }{15} )=Cos \frac{ \pi }{3}= \frac{1}{2}$

$Cos \frac{3 \pi }{8} Sin \frac{5 \pi }{24}-Cos \frac{5 \pi }{24}Sin \frac{3 \pi }{8}=Sin( \frac{5 \pi }{24}- \frac{3 \pi }{8})=Sin \frac{5 \pi -9 \pi }{24}=-Sin \frac{4 \pi }{24}=$$-Sin \frac{ \pi }{6}=- \frac{1}{2}$

$\frac{Sin7Cos47+Cos7Sin47}{Sin13Cos41+Cos13Sin41}= \frac{Sin(7+47)}{Sin(13+41)}= \frac{Sin54}{Sin54}=1\\\\ \frac{Cos51Cos12-Sin51Sin12}{Sin13Cos14+Cos13Sin14}= \frac{Cos(51+12)}{Sin(13+14)}= \frac{Cos63}{Sin27} = \frac{Cos(90-27)}{Sin27}= \frac{Sin27}{Sin27} =1$

Sin105Sin75=Sin(180 - 75) * Sin75 = Sin75 * Sin75 = Sin²75

$Sin105 + Sin15 =2Sin \frac{105+15}{2} Cos \frac{105-15}{2}=2Sin60Cos45= 2\frac{ \sqrt{3} }{2}* \frac{ \sqrt{2} }{2}$$= \frac{ \sqrt{6} }{2}\\\\Cos165+Cos75=2Cos \frac{165+75}{2} Cos \frac{165-75}{2}=2Cos120Cos45=2*(- \frac{1}{2}*$$\frac{ \sqrt{2} }{2} =- \frac{ \sqrt{2} }{2}$