Question!

**BrownianMan** · Nov25-09, 09:50 PM

There's no answer for this in the back of the book and I just want to make sure I did it correctly. If anyone could post their answer, I would appreciate it! Thanks.

Cos^2(x) + sin^2 (x) =1 for every real value of x.(Pythagorean theorem)
What real values of x will be a solution to cos^n (x)- sin^n (x) =1, for a given positive integer n?

**Mark44** · Nov25-09, 10:33 PM

What values did you get?

**BrownianMan** · Nov25-09, 10:36 PM

x=0.64 + pi/4 +/- 2*pi*n, 0.64 + 3pi/4 +/- 2*pi*n

**Mark44** · Nov26-09, 02:08 AM

Can you show me how you got those numbers? BTW, they are not correct.

**Mentallic** · Nov26-09, 02:11 AM

What's interesting is that the solutions to x are the same for all odd, and the other set of solutions are equal for all even n.
Your answers are incorrect. Just test them and you'll see.

There is one solution for x that works with all positive n, and that is x=0.

**Mark44** · Nov26-09, 10:46 AM

Originally Posted by Mentallic

What's interesting is that the solutions to x are the same for all odd, and the other set of solutions are equal for all even n.

A more accurate way to say this is that there is one set of solutions for all odd integers n, and another set of solutions for all even integers n.

**BrownianMan** · Nov26-09, 12:43 PM

Ok, so if n is even x=0 x=k*pi, and if n is odd x=0 x=3pi/2+2*k*pi???

**Mark44** · Nov26-09, 04:26 PM

Originally Posted by BrownianMan

Ok, so if n is even x=0 x=k*pi, and if n is odd x=0 x=3pi/2+2*k*pi???

For n even, yes. For n odd, what you have is correct, there is another bunch of solutions. What I did was draw graphs of y = cos^n(x) and y = sin^n(x) + 1, and found the points where the two graphs intersect. The graphs of odd powers of sine and cosine look pretty much like the graphs of sine and cosine.

**Mentallic** · Nov26-09, 10:30 PM

Originally Posted by Mark44

A more accurate way to say this is that there is one set of solutions for all odd integers n, and another set of solutions for all even integers n.

Yes, thankyou

I'm stepping into fields I have yet to learn so my terminology would be questionable at best. I just hope that I can get the point across.