math question!

[hassam]

If n ∈ N, solve the equation cos*n x − sin*n x = 1, for x ∈ R.
*n means exponent
∈ means member of

 

[MAVtKnmJ]

Think about this.

If x = 0, then the solution is n ∈ mathbbZ^+

Hope that helps

P.S

Forgot to give details ...

If for x= 0, cos(x)= cos(0)= 1, sin(x)= sin(0)= 0 and the equation is true for n any positive integer (actually it is true for any integer).

But the problem here is to determine those positive integers, n, such that the equation is true for ALL x. More to the point would be to look at x= pi/4 for which cos(x)= sin(x)

 

[usman]

@ actually it is true for any integer: i am afraid this ain't the case cos:
{sin(x)}^n = 0^-1 (when n = -1 and x = 0) which is undefined, considering that 0 is also an integer

 

[Zishi]

If n ∈ N, solve the equation cos*n x − sin*n x = 1, for x ∈ R.
*n means exponent
∈ means member of

From where did you get this question? Never solved these kinda questions... o_o

 

[anzaar]

it can be solved by this method.
for n=1
it is true.
let it is true for n=k
you have to prove it is also true for n=k+1
hence it is true for all real numbers

 

[hassam]

NMO---National math olympiad----thought dat sum 1 myt help me out

 

[hassam]

ya anzaar u talkin abt mathematical induction ryt....? gUd suPer idea

 

[OakMoon!]

I found the same question. Didnt get a clue of what is it talking about :S

 

[hassam]

difficult question....hmm....

 

[usman]

Forn n ≥ 2 we have:-
1 = cos^n(x) - sin^n(x) ≤ Abs [cos^n(x) - sin^n(x)] ≤ Abs [cos^n(x)] + Abs [sin^n(x)] ≤ cos^2(x) + sin^2(x) = 1

Hence sin^2(x) = Abs [sin^n(x)] and cos^2(x) = Abs [cos^n(x)], from which it follows that sin(x), cos(x) ∈ {1,0,-1} ---------> x ∈ pi*Z/2. By inspection one obtains the set of solutions
{m*pi, m ∈ Z} for even n and {2m*pi, 2m*pi - pi/2, m ∈ Z} for odd n.

For n = 1 we have 1 = cos(x) - sin(x) = -sqrt(2)*sin(x - pi/4), which yields the set of solutions
{2m*pi, 2m*pi - pi/2, m ∈ Z}.

Hope that helps..!!

 

[MAVtKnmJ]

Do you know how old this thread is? Plus it was already solved.