Checking this answer regarding a trig problem

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SUMMARY

The discussion centers on the trigonometric problem of proving the existence of constants C and ∝ such that Ccos(x+∝) = Asinx + Bcosx, specifically for A = B = 1. The author presents the solution C = √2 and ∝ = -π/4. A participant questions the validity of an alternative solution using C = -√2 and ∝ = -3π/4, which does not satisfy the condition C ≥ 0. The consensus confirms that only the original solution is valid.

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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?
 
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rxh140630 said:
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##?
 
rxh140630 said:
... with C ≥ 0 ...
...
but plugging C = - \sqrt{2},
 

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