Auxiliary Angles (1 Viewer)

[ihatehippos101]

for this question, are you supposed to change the domain to pi/6 ≤ x + pi/6 ≤ 13pi/6 ?
since the new form is sqrt8*sin(x+pi/6)
i remember faintly that some worked solutions had that but others don't for different auxiliary angle questions.

so my solutions would be 7pi/12 and 25pi/15 while the worked solutions are pi/12 and 7pi/12, but with the new domain pi/12 would be out of it.

 

[alphxreturns]

View attachment 48559
for this question, are you supposed to change the domain to pi/6 ≤ x + pi/6 ≤ 13pi/6 ?
since the new form is sqrt8*sin(x+pi/6)
i remember faintly that some worked solutions had that but others don't for different auxiliary angle questions.

so my solutions would be 7pi/12 and 25pi/15 while the worked solutions are pi/12 and 7pi/12, but with the new domain pi/12 would be out of it.

yes, u change the domain but only to solve for x + π/6 = π/4, 3π/4 (this has domain like you said) (π/4 would still be valid as 1/4 > 1/6)


but then u have to change the domain back (subtract all by π/6) to solve for x, meaning the domain returns to normal [0,2π]

 

[Socialism]

what subject is this? I assume advanced?
i think we've done questions that look a little like that? so probably?

 

[coolcat6778]

what subject is this? I assume advanced?
i think we've done questions that look a little like that? so probably?

extension 1. u learnt sin(a+b) = sinacosb + cosasinb?

 

[Socialism]

extension 1. u learnt sin(a+b) = sinacosb + cosasinb?

OHHH ofc... @anarcho-monarchism was showing me some Q's from ext that must be where I've seen that