(tan15∘)x(tan25∘)x(tan35∘)x(tan85∘)
I know the answer but I do not know the solution
answer: 1
help me pls
+26396 (tan15∘)x(tan25∘)x(tan35∘)x(tan85∘)
Formula:
sin(x)sin(y)=12(cos(x−y)−cos(x+y))cos(x)cos(y)=12(cos(x−y)+cos(x+y))sin(2x)=2⋅sin(x)⋅cos(x)
tan(15∘)⋅tan(25∘)⋅tan(35∘)⋅tan(85∘)=sin(15∘)⋅sin(25∘)⋅sin(35∘)⋅sin(85∘)cos(15∘)⋅cos(25∘)⋅cos(35∘)⋅cos(85∘)=[sin(85∘)⋅sin(15∘)]⋅[sin(35∘)⋅sin(25∘)][cos(85∘)⋅cos(15∘)]⋅[cos(35∘)⋅cos(25∘)]sin(85∘)⋅sin(15∘)=12(cos(85∘−15∘)−cos(85∘+15∘))sin(85∘)⋅sin(15∘)=12(cos(70∘)−cos(100∘))sin(35∘)⋅sin(25∘)=12(cos(35∘−25∘)−cos(35∘+25∘))sin(35∘)⋅sin(25∘)=12(cos(10∘)−cos(60∘))cos(85∘)⋅cos(15∘)=12(cos(85∘−15∘)+cos(85∘+15∘))cos(85∘)⋅cos(15∘)=12(cos(70∘)+cos(100∘))cos(35∘)⋅cos(25∘)=12(cos(35∘−25∘)+cos(35∘+25∘))cos(35∘)⋅cos(25∘)=12(cos(10∘)+cos(60∘))=12(cos(70∘)−cos(100∘))⋅12(cos(10∘)−cos(60∘))12(cos(70∘)+cos(100∘))⋅12(cos(10∘)+cos(60∘))=(cos(70∘)−cos(100∘))⋅(cos(10∘)−cos(60∘))(cos(70∘)+cos(100∘))⋅(cos(10∘)+cos(60∘))|cos(100∘)=cos(90∘+10∘)=−sin(10∘)=(cos(70∘)+sin(10∘))⋅(cos(10∘)−cos(60∘))(cos(70∘)−sin(10∘))⋅(cos(10∘)+cos(60∘))|cos(70∘)=cos(90∘−20∘)=sin(20∘)=(sin(20∘)+sin(10∘))⋅(cos(10∘)−cos(60∘))(sin(20∘)−sin(10∘))⋅(cos(10∘)+cos(60∘))|cos(60∘)=12=(sin(20∘)+sin(10∘))⋅(cos(10∘)−12)(sin(20∘)−sin(10∘))⋅(cos(10∘)+12)=sin(20∘)⋅cos(10∘)−12⋅sin(20∘)+sin(10∘)⋅cos(10∘)−12⋅sin(10∘)sin(20∘)⋅cos(10∘)+12⋅sin(20∘)−sin(10∘)⋅cos(10∘)−12⋅sin(10∘)|sin(10∘)⋅cos(10∘)=12⋅sin(20∘)=sin(20∘)⋅cos(10∘)−12⋅sin(20∘)+12⋅sin(20∘)−12⋅sin(10∘)sin(20∘)⋅cos(10∘)+12⋅sin(20∘)−12⋅sin(20∘)−12⋅sin(10∘)|12⋅sin(20∘)−12⋅sin(20∘)=0=sin(20∘)⋅cos(10∘)+0−12⋅sin(10∘)sin(20∘)⋅cos(10∘)+0−12⋅sin(10∘)=sin(20∘)⋅cos(10∘)−12⋅sin(10∘)sin(20∘)⋅cos(10∘)−12⋅sin(10∘)=1
+118703 Thanks Heureka :)
I have only just started working through this.
I am sure it is very clear and I will have no problem following it.
However, I would not have been able to put that string of logic together myself.
Maybe as I work through it some light may come on for me,
I hope so :))
