8.aux.27 Simplify the trig expression

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Discussion Overview

The discussion revolves around the simplification of the trigonometric expression $\dfrac{{\cos 2x}}{{\cos x - \sin x}}$. Participants explore the steps taken to simplify the expression and address potential issues related to the conditions under which the simplification holds.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents a simplification of the expression, arriving at $\cos x + \sin x$.
  • Another participant confirms the correctness of the simplification.
  • A third participant notes that the simplification is valid only under the condition that $\cos(x) \neq \sin(x)$, highlighting the importance of this condition to avoid an indeterminate form.
  • A later reply agrees with the previous point, emphasizing that without this condition, the expression could lead to a $0/0$ scenario.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the simplification, but there is a clear acknowledgment of the need for specific conditions to avoid undefined behavior in the expression.

Contextual Notes

The discussion highlights the importance of stating conditions under which the simplification is valid, particularly the requirement that $\cos(x) \neq \sin(x)$ to prevent division by zero.

karush
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$\tiny{8.aux.27}$
Simplify the expression
$\dfrac{{\cos 2x\ }}{{\cos x-{\sin x\ }\ }}
=\dfrac{{{\cos}^2 x-{{\sin}^2 x\ }\ }}{{\cos x\ }-{\sin x\ }}
=\dfrac{({\cos x}-{\sin x})({\cos x}+{\sin x\ })}{{\cos x}-{\sin x}}
=\cos x +\sin x$

ok spent an hour just to get this and still not sure
suggestions?
 
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it's correct
 
With the proviso that we only use x s.t. [math]cos(x) \neq sin(x)[/math]. Since the reason for this has left the expression we need to state that.

-Dan
 
good point otherwise you get 0/0
 

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