(sin x)^n-(cos x)^n =1 for all x

[dpsguy]

Find the value of n such that

(sin x)^n-(cos x)^n =1 for all x

I really have no idea how to go about it,or even whether such a number exists.Anybody willing to help?

 

[Tide]

Obvious solutions exist where sin x = 1 and cos x = 0 and where sin x = 0 and cos x = 1. You should be able to find others! :)

 

[Galileo]

If it is supposed to hold for ALL x. It should in particular hold for x=0, right?

 

[Werg22]

sin(x)^2 + cos(x)^2=1

sin(x)^2=1-cos(x)^2

and

sin(x)^n - cos(x)^n=1

sin(x)^n=1 + cos(x)^n

Let n=2l

sin(x)^2l=1 + cos(x)^2l

And

sin(x)^2l=(1-cos(x)^2)^l

So

1 + cos(x)^2l=(1-cos(x)^2)^l

With binominal theorem find a value such as,

(1-b^2)^l=1 + b^2l

 

[Werg22]

That is impossible. cos(x)<1 and sin(x)<1. A value inferior to one raised to any power is always inferior to 1. I think you mean cos(x)^n + sin(x)^n=1. In that case it's pretty simple; first let's proove that n is even;

sin(180)=0
cos(180)=-1

-1^n=1, thus n is even.

Let n=2l

Let a right triangle with the angle x have sides a,b and c such as c^2=a^2 + b^2.

Sin(x)^2=a^2/c^2
Cos(x)^2=b^2/c^2

(b^2/c^2)^l + (a^2/c^2)^l=1

a^2l + b^2l/c^2l=1

Since a^2 + b^2=c^2

a^2l + b^2l/(a^2 + b^2)^l=1
a^2l + b^2l=(a^2 + b^2)^l

By the binominal theorem we know that the only possible value for l; such as

(a+b)^l=a^l+b^l

is 1.

Thus l=1

Since n=2l
n=2

 

[dpsguy]

How about n<0?
Surely (cosecx)^n-(secx)^n can be 1 for some n when x is not equal to a(pi/2) where a=0,1,2,... ,not for ALL x.

 

[dpsguy]

How about for all x where cosecx>secx?