Question #c2e32

Question

1564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-20T09:31:25

Considering the ratio

$\frac{sin 8 ϕ - sin 10 ϕ}{cos 10 ϕ - cos 8 ϕ}$ $(sin8phi-sin10phi)/(cos10phi-cos8phi)$

Using following identities to simplify the above ratio

$sin c - sin d = 2 cos (\frac{c + d}{2}) sin (\frac{c - d}{2})$ $color(red)(sinc-sind=2cos((c+d)/2)sin((c-d)/2))$

and

$cos c - cos d = 2 sin (\frac{c + d}{2}) sin (\frac{d - c}{2})$ $color(red)(cosc-cosd=2sin((c+d)/2)sin((d-c)/2))$

Now the ratio

$\frac{sin 8 ϕ - sin 10 ϕ}{cos 10 ϕ - cos 8 ϕ}$ $(sin8phi-sin10phi)/(cos10phi-cos8phi)$

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

$=(cos9phi)/(sin9phi)=cot9phi$

So we can write

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

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

proved