A few qs. (1 Viewer)

[香港!]

Could somebody help me with these questions:

1. Prove that (1 + sin2x)/(1 - sin2x) = tan²(x + pi/4)

2. File attached - (2unit?) Maxmin question
I can get x = ac/(a + b) after some messy algebra but can't show it's the shortest path... also please do show any better methods to approach it...

thanks.

"1. Prove that (1 + sin2x)/(1 - sin2x) = tan²(x + pi/4)"
LHS=(1+sin2x)\(1-sin2x)
=(1+2sinxcosx)\(1-2sinxcosx)
=((1\cos²x)+(2sinxcosx\cos²x)) \ ((1\cos²x)-(2sinxcosx\cos²x))
=(sec²x+2tanx)\(sec²x-2tanx)
=(tan²x+2tanx+1)\(tan²x-2tanx+1)
=(tanx+1)²\(tanx-1)²
=[(tanx+1)\(tanx-1)]²
RHS=tan²(x+pi\4)
=[tan(x+pi\4)]²
=[(tanx+tan(pi\4) )\1-tanx tan (pi\4) ]²
=[(tanx+1)\(1-tanx)]²
=[(tanx+1)\(tanx-1)]² (since it's squared the i can switch it)
.: LHS=RHS

 

[香港!]

aye y does my post go on top LoL

 

[word.]

Could somebody help me with these questions:

1. Prove that (1 + sin2x)/(1 - sin2x) = tan²(x + pi/4)

2. File attached - (2unit?) Maxmin question
I can get x = ac/(a + b) after some messy algebra but can't show it's the shortest path... also please do show any better methods to approach it...

thanks.

 

[Trev]

Q2b)
when x=ac/(a+b).
Distance DC = ac/(a+b)
Distance CE = c - ac/(a+b) = (ac+bc-ac)/(a+b) = bc/(a+b)
Therefore the angles alpha and beta are:
tanα = (opp/hyp) = (a/[ac/(a+b)]) = (a+b)/c
angle α = tan^-1[(a+b)/c]
tanβ = (opp/hyp) = (b/[bc/(a+b)]) = (a+b)/c
angle β = tan^-1[(a+b)/c]
Hence shortest path occurs when α = β = tan^-1[(a+b)/c]

 

[Stefano]

香港!
Member

HSC: 2010


WTF??? You serious ?

 

[acmilan]

香港!
Member

HSC: 2010


WTF??? You serious ?

Give or take 5 years, mainly take

 

[Stefano]

Haha, cool!
It was a bit unerving :S