Page 1 of 1

Cofunction problem

Posted: Wed Dec 14, 2005 3:35 pm
by Ambush Commander
Recently, we learned a new technique for solving otherwise hairy trigonometric equations. It involved substiting Sin[x] with Cos[Pi/2-x] and all variations, and then changing Sin[x] == Sin[y] into x == y + 2Pi*k where k is an integer and 2Pi is the period of the function.

I've run into trouble applying this technique on this equation: Sin[x]+Cos[x]==0. I know that the answer is derivable by moving the cosine on one side of the equation, squaring both sides, and using the Pythagorean Identity Sin^2[x] + Cos^2[x] == 1.

However, applying the cofunction technique yields this:

Code: Select all

Sin[x]+Cos[x]==0
Sin[x]==-Cos[x]
Sin[x]==-Sin[Pi/2-x]
Sin[x]==Sin[x-Pi/2]
x==x-Pi/2+2Pi*k
Pi/2==2Pi*k
Now, somehow the x's canceled out, but there is only one solution for k! That has totally confused me... have I messed up in my steps? What did I miss?

Posted: Wed Dec 14, 2005 5:51 pm
by RobertGonzalez
Would Cos[x] = Sin[Pi/2-x]? I vaguely remember something about negative inverses or reciprocals when converting Sin's and Cos's. Of course that was 13 years and five kids ago, but there is something about (Sin[x] = Cos[Pi/2-x]) == (Cos[x] = Sin[Pi/2-x]) that looks a little strange.

Like I said, it has been a long time since trig, calculus, linear algebra and differential equations. However, now I can say that I actually responded to a Ambush Commander question of this magnitude. Cool.

Posted: Wed Dec 14, 2005 9:41 pm
by Ambush Commander
Hmm... I fairly certain that is the cofunction... if not... well all I can say is "Stoopid!" (like those little worms do when you inflict-self damage)

::Ponders it::

No... I think the cofunction is right. Cos[x] = Sin[Pi/2-x] is equiavlent to Cos[x] = Sin[90-x], and the cofunctions delve on the fact that Cos[O] = x/r and Sin[O] = y/r, and taking the difference of 90 degrees and the original angle swaps the x and the y.
However, now I can say that I actually responded to a Ambush Commander question of this magnitude. Cool.
::Is flattered:: :oops: