Trig Question - HSC Exam Style 3U (1 Viewer)

[TooSleepy]

Not exactly sure where this came from originally but it's on a sheet our teacher gave us And....i have no idea how to do it:

Given that 0 < x < π/4 , prove that

tan(π/4 + x) = ( cosx + sinx )/( cosx - sinx )

 

[Aesytic]

LHS = tan(pi/4 + x)
= [tan(pi/4) + tanx]/[1-(tan(pi/4)*tanx)]
= [1 + tanx]/[1 - tanx] since tan(pi/4) = 1
= [1 + sinx/cosx]/[1-sinx/cosx]
multiplying top and bottom by cosx,
= [cosx + sinx]/[cosx - sinx]
= RHS

 

[TooSleepy]

Ohhh so using the compound angle thing....
i get it!
Thanks aesytic!!

 

[Carrotsticks]

 

[TooSleepy]

Oh thanks too carrotsticks!
Two ways to do it....hmm, i wonder which one's better.....
both get you the marks though so i guess it doesn't really matter

 

[ahdil33]

Same thing really isn't it? But I think the first is easier because expanding a compound angle is easier than realising it could be put in the form of one, especially as tan (pi/4) = 1

 

[Carrotsticks]

Oh thanks too carrotsticks!
Two ways to do it....hmm, i wonder which one's better.....
both get you the marks though so i guess it doesn't really matter

Well Aeystic proved one direction (baby you light up my world like nobody else) whereas I did another direction, but his one is more intuitive than mine. For some reason I read the question as prove (expression with sin and cos) = (expression with tan) when you in fact had written it the other way around!

 

[Sanjeet]

Why do they give you 0 < x < π/4 as a condition if it is not needed in the solution?

 

[Sy123]

Why do they give you 0 < x < π/4 as a condition if it is not needed in the solution?

Carrotsticks cancelled out cos x, so in the given domain, he can because cos x =/= in that domain. (I think)

EDIT: I graphed both sides of the equation on Geogebra, and they seem to be the exact same for all values of x, I was guessing that they might of given you that domain, because it was only exactly the same for that domain, (or in the domain similar to it)