trig help (1 Viewer)

[chewy123]

prove



1+cos A+sin A 1+cosA
__________ ________
1-cosA+sinA = sinA


thx

 

[lyounamu]

prove



1+cos A+sin A 1+cosA
1-cosA+sinA = sinA

thx

Your question is poorly written. Can you write it again?

 

[chewy123]

(1+ cosA + sinA) / (1 - cosA + sinA) = (1+cosA) / sinA

 

[lyounamu]

(1+ cosA + sinA) / (1 - cosA + sinA) = (1+cosA) / sinA

Whew, got it. Solution is to come here:

(1+cosA)/sinA . (1/sinA - cosA/sina + 1)/(1/sinA - cosA/sinA + 1)
= (1/sinA - cosA/sinA + 1 + cosA/sinA - cos^2(A))/sinA + cosA/(1-cosA+sinA)
= (times everything by sin A on top and bottom)
= (1 - cosA + sinA + cosA - cos^2(A) + cosAsinA)/(sinA-cosAsinA+sin^2(A))
= 1-cos^2(A) + cosAsinA + sinA / (sinA-cosAsinA + sin^2(A))
= (sin^2(A) + cosAsinA +sinA) /(sinA-cosAsinA + sin^2(A))
= (sinA + cosA + 1)/(1-cosA +sinA)

I went from RHS = LHS and this is still acceptable as this is "prove" question. If it was "show" question, you must start from LHS to RHS.

 

[lyounamu]

You do 3U maths so you know t-relationships right?

I substituted sinA=2t/(1+t^2) and cosA=(1-t^2)/(1+t^2), simplified and got LHS=RHS=1/t

I don't think this is 3 unit question.

You don't have to use 3 unit method...I mean I tried not to..

 

[lyounamu]

xD t relationships was the first thing that came into my head. It looks so complicated so I had no idea how to do it otherwise lol

This question is little complicated yes. But I went against using that method because the OP doesn't do 3 Unit which means that he didn't learn t-method.

 

[chewy123]

thx all, may you all succeed in the hsc so long you're slight below meXDXD
just kidding