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

  • Thread starter dpsguy
  • Start date
In summary, the value of n such that (sin x)^n-(cos x)^n =1 for all x is 2. This can be proven through the binomial theorem and the properties of right triangles. However, it is not possible for the equation to hold for all values of x when n is negative or when x is equal to a(pi/2) where a is a non-negative integer.
  • #1
dpsguy
69
0
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?
 
Physics news on Phys.org
  • #2
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! :)
 
  • #3
If it is supposed to hold for ALL x. It should in particular hold for x=0, right?
 
  • #4
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
 
Last edited:
  • #5
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
 
Last edited:
  • #6
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.
 
  • #7
How about for all x where cosecx>secx?
 

1. What is the equation (sin x)^n-(cos x)^n=1 for all x?

This equation is a trigonometric identity that states that for any value of x, the nth power of the sine of x minus the nth power of the cosine of x will always equal 1.

2. How do we prove that (sin x)^n-(cos x)^n=1 for all x?

This identity can be proven using mathematical induction or by using the power-reducing formulas for sine and cosine.

3. What does the variable n represent in the equation (sin x)^n-(cos x)^n=1 for all x?

The variable n represents the power that the sine and cosine functions are raised to.

4. Can (sin x)^n-(cos x)^n=1 for all values of x?

Yes, this equation holds true for all values of x, including positive and negative numbers, decimals, and even complex numbers.

5. How is the equation (sin x)^n-(cos x)^n=1 for all x used in mathematics?

This equation is commonly used in trigonometry and calculus to simplify expressions and solve problems involving trigonometric functions. It is also used in physics and engineering to describe and analyze periodic motion and oscillations.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
369
  • Introductory Physics Homework Help
Replies
28
Views
270
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
1
Views
83
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
193
  • Introductory Physics Homework Help
Replies
2
Views
186
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
623
  • Introductory Physics Homework Help
Replies
29
Views
810
Back
Top