tanα:tanβ=x:y

Question

show that sin(α-β)=(x-y)/(x+y)* sinθ where α+β=θ

Answer 1

For alpha I've used a, for beta I've used b, and for theta I've used ø.

sin(a-b)=sin(a)cos(b)-cos(a)sin(b). Let tan(a)=X and tan(b)=Y then sin(a)=X/√(1+X^2), cos(a)=1/√(1+X^2), sin(b)=Y/√(1+Y^2), cos(b)=1/√(1+Y^2) and tan(a)/tan(b)=X/Y. So x=X and y=Y.

sin(a-b)=x/√((1+x^2)(1+y^2))-y/√((1+x^2)(1+y^2))=(x-y)/√((1+x^2)(1+y^2)).

But:

sinø=sin(a+b)=x/√((1+x^2)(1+y^2))+y/√((1+x^2)(1+y^2))=

(x+y)/√((1+x^2)(1+y^2)).

Therefore sin(a-b)/sin(a+b)=sin(a-b)/sinø=(x-y)/(x+y);

sin(a-b)=((x-y)/(x+y))*sinø QED.

 

Comments

you beat me by 57 secs !!! :)

Answer 2

show that sin(α-β)=(x-y)/(x+y)* sinθ where α+β=θ

 

sin(α-β)/sinθ =(x-y)/(x+y) where θ​=α+β

 

​sin(α-β)/sin(α+β) =

{sin(α).cos(β) - cos(α).sin(β)} / {sin(α).cos(β) + cos(α).sin(β)} =

dividinbg above and below by sin(α).cos(β),

{1 - cot(α).tan(β)} / {1 + cot(α).tan(β)} =

{1 - tan(β)/tan(α)} / {1 + tan(β)/tan(α)} =

​using tan(α)/tan(β) = x/y,

{1 - y/x) / {1 + y/x} =

​{x - y} / {x + y}