Question #c2e32

Question

1377 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-20T09:31:25

Considering the ratio

#(sin8phi-sin10phi)/(cos10phi-cos8phi)#

Using following identities to simplify the above ratio

#color(red)(sinc-sind=2cos((c+d)/2)sin((c-d)/2))#

and

#color(red)(cosc-cosd=2sin((c+d)/2)sin((d-c)/2))#

Now the ratio

#(sin8phi-sin10phi)/(cos10phi-cos8phi)#

#=(cancel2cos((8phi+10phi)/2)cancelsin((8phi-10phi)/2))/(cancel2sin((10phi+8phi)/2)cancelsin((8phi-10phi)/2))#

#=(cos9phi)/(sin9phi)=cot9phi#

So we can write

#(sin8phi-sin10phi)/(cos10phi-cos8phi)=cot9phi#

#=>(sin8phi-sin10phi)=cot9phi(cos10phi-cos8phi)#

proved