Checking this answer regarding a trig problem

[rxh140630]

Homework Statement:

Prove that if A, B are given real numbers there exists C and ∝ with C ≥ 0 such that Ccos(x+∝) = Asinx + Bcosx. Determine C and ∝ if A=B=1

Relevant Equations:

cos(x+y)= cosxcosy-sinxsiny

Author gave solution $C = \sqrt{2}, ∝ = -pi/4$

but plugging $C = - \sqrt{2}, ∝ = -3pi/4$ into cos(x+y) and leaving the x I get $\sqrt{2}Cos(x+3pi/4) = sinx+cosx$

Is my answer valid as well?

 

[PeroK]

Homework Statement:: Prove that if A, B are given real numbers there exists C and ∝ with C ≥ 0 such that Ccos(x+∝) = Asinx + Bcosx. Determine C and ∝ if A=B=1
Relevant Equations:: cos(x+y)= cosxcosy-sinxsiny

Author gave solution $C = \sqrt{2}, ∝ = -pi/4$

but plugging $C = - \sqrt{2}, ∝ = -3pi/4$ into cos(x+y) and leaving the x I get $\sqrt{2}Cos(x+3pi/4) = sinx+cosx$

Is my answer valid as well?

What does your answer give at $x = 0$?

 

[haruspex]

... with C ≥ 0 ...
...
but plugging $C = - \sqrt{2},$