trig (1 Viewer)

[twistedrebel]

(i) SinA + Sin2A / 1 + CosA + CoS2A = Tan A

(t-method)
Sin2x (1 - cosx) / cosx (1-cos2x) = t

 

[bleaver]

Are those two separate questions?
I hope so..haha.

Here's the first:
LHS= (SinA+2SinACosA)/(1+cosA+cos^2A-sin^A)
=(SinA+2SinAcosA)/(1+cosA+cos^A-1+cos^2A)
=SinA/CosA * (2CosA+1)/(2CosA+1) (Factorising the last line)
=TanA
=RHS


Brb and ill do part two. Providing thats a separate question..

 

[bleaver]

Ok part 2.

LHS=(sin2x)(1-cosx)/(cosx)(1-cos2x)
=(2sinxcosx)(1-cosx)/(cosx)(1-cos^2x+sin^2x)
=(2sinx)(1-cosx)/(1-cos^2x+sin^2x)
=(2sinx)(1-cosx)/(2sin^2x)
=(1-cosx)/(sinx)

So far I've just used sin2x, cos2x, 1-cos^2x substitutions to simplify.

Now substituting cosx=(1-t^2)/(1+t^2), and sinx=2t/(1+t^2)
(1-cosx)/(sinx)
=(1-((1-t^2)/(1+t^2)))/((2t)/(1+t^2))
=2t^2/2t
=t
=RHS

Sorry that gets a bit messy. Still working out how to use latex.

 

[bleaver]

Sorry, just managed to get LaTeX working:

Part A:



Part B:



So far only using the following substitutions to simplify:


Finally: