Cofunction problem
Posted: Wed Dec 14, 2005 3:35 pm
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:
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?
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