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How to prove if the next number is rational, without using trigonometric table and knowing that tg a=3/5?
A=(17/15)*[sin(2a)]+(17/8)*[cos(2a)]+ctg80*ctg20*ctg40
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4515*{cos9*tan78}*4567766/87
tana = 3/5. sina = 3/sqrt(3^2+5^2)= 3/sqrt34. cosa=5/sqrt34
sin2a=2sinacosa=2*3*5/34 =30/34. cos2A= 2cos^2a-1=2*5^2/34 -1 = 16/34
cot80*ctg20*ctg40=tan10*cot20*tam50 as ctg theta=tan(90-theta).
tan(30-20)tan(30+20)/tan20
={[(tan30)^2-(tan20)^2]/[1-(tan30*tan20)^2]}1/tan20
={[1/3-(tan20)^2]/[1-(1/3)(tan20)^2]]/tan20
={[1-3(tan20)^2]/[3-(tan20)^2]}/(tan20)
={1-3(tan20)^2]cos20/[3-(tan20)^2]sin20
=[(cos20)^2-3(sin20)^2}cos20/{[3(cos20)^2-(sin20)^2]sin20]
={(cos20)^3-3(1-(cos20)^20*cos20)}/{[3(1-(sin20)^20-(sin20)^2]sin20}
=[4(cos20)^3-3cos20]/[3sin20-4(sin20)^3]
=cos (3*20)/sin(3*20). as cos3x=4(cosx)^3-3cosx and Sin3x= 3sinx-4(sinx)^3
=ctg 60
=1/sqrt3.
Therefore,
A=(17/15)(30/34) + 17/8(16/34)+1/sqt3
=1+ 1 +1/sqr3
2+1/sqrt3 =2.577350269
=( 2sqrt3+1)/sqrt3
=(6+sqrt3)3
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