Trig Proving (1 Viewer)

RivalryofTroll · Mar 29, 2012

Given:
tanx = ktany

Prove/Show that:
(k+1)sin(x-y) = (k-1)sin(x+y)

qrpw · Mar 29, 2012

My method:

$tanx = ktany$

$\frac{sinx}{cosx} = k\frac{siny}{cosy}$

$\frac{sinxcosy}{cosxsiny} = k$

You require sin(x+y)'s in the answer, so use

$\frac{a+b}{a-b} = \frac{c+d}{c-d} where \frac{a}{b} = \frac{c}{d}$

$\frac{sinxcosy + cosxsiny}{sinxcosy - cosxsiny} = \frac{k+1}{k-1}$

$\frac{sin(x+y)}{sin(x-y)} = \frac{k+1}{k-1}$

$(k+1)sin(x-y) = (k-1)sin(x+y)$

$Q.E.D$

math man · Apr 5, 2012

we are given:
$tanx=ktany$
the answer has sin only so lets change evrything in terms of sin and cos
$\frac{sinx}{cosx}=k\frac{siny}{cosy} \\ sinxcosy=ksinycosx \\ \frac{1}{2}(sin(x+y)+sin(x-y)=\frac{k}{2}(-sin(x-y)+sin(x+y)) \\ sin(x+y) + sin(x-y) = -ksin(x-y) + ksin(x+y) \\ (k+1)sin(x-y)=(k-1)sin(x+y)$
in there i used the compound expansion formula as i noticed i has sinxcosy which is one half of each of sin(x-y) and sin(x+y) and knew how to manipulate it from there

Trig Proving (1 Viewer)

RivalryofTroll

Sleep Deprived Entity

qrpw

Member

math man

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)