Help

[davidle_10]

Anyone care to help me solve this?
If cos(α)=sin(α-β)sin(β), prove that tan(α-β)tan(β)=1/2

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?

edit: stupid me, it's an identity not a function

[pi]

I tried, but gave up when the algebra got too messy in tan(a-B)...

Maybe I'll try later with a calc for easier simplifications.

(sorry)

[pi]

It is really ugly, so I'm just going to explain what to do

If cos(α)=sin(α-β)sin(β), prove that tan(α-β)tan(β)=1/2

1. Expand sin(a-b)       <-- should know this identity
2. Rearrange and solve for sin(a) and cos(b)
3. Solve for sin(b) by letting sin(b)=x and solve as a quadratic
4. Find tan(a) using sin(a)/cos(a)
5. Find tan(b) using sin(b)/cos(b)
6. Hence, find tan(a-b)         <-- should know this identity
7. Multiply tan(a-b) and tan(b), should get 1/2, hence the proof

Its ugly, especially after step 2 (ie. most of it). Use a graphics calc for ease

[davidle_10]

I figured it out using an easier way but thank you for your contribution anyway.
What I did was this:
cos(a)=cos((a-b)+b)
cos((a-b)+b)=cos(a-b)cos(b)-sin(a-b)sin(b)
since cos(a)=cos((a-b)+b), sin(a-b)sin(b)=cos(a-b)cos(b)-sin(a-b)sin(b)
2sin(a-b)sin(b)=cos(a-b)cos(b)
sin(a-b)sin(b)/cos(a-b)cos(b)=1/2
tan(a-b)tan(b)=1/2

[pi]

I figured it out using an easier way but thank you for your contribution anyway.
What I did was this:
cos(a)=cos((a-b)+b)
cos((a-b)+b)=cos(a-b)cos(b)-sin(a-b)sin(b)
since cos(a)=cos((a-b)+b), sin(a-b)sin(b)=cos(a-b)cos(b)-sin(a-b)sin(b)
2sin(a-b)sin(b)=cos(a-b)cos(b)
sin(a-b)sin(b)/cos(a-b)cos(b)=1/2
tan(a-b)tan(b)=1/2

Shit thats a good way!