Prove that the inequality sin^8 (x)+cos^10 (x)

It is not true that sin^8(x) + cos^10(x) is always less than 1.

For example when x = 0, sin(x) = 0 and cos(x) = 1

Therefore: sin^8(x) + cos^10(x) = 0^8 + 1^10 = 0 +1 = 1

However we can prove that:

sin^8(x) + cos^10(x) is less than or equal to one.

The proof is as follows.

We know that

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

This equality has three general relative values of sin^2(x) and cos^2(x).

1. sin^2(x) = 1 and cos^2(x) = 0


2. sin^2(x) = 0 and cos^2(x) = 1


3. Both sin^2(x) and cos^2(x) have a value between 0 and one.

Under the first two condition the sin^8(x) + cos^10(x) = 1 because any exponent of 1 = 1 and. any exponent of 0 = 0.

Under the third condition let us say sin^2(x) = A, and cos^2(x) = B

Then: sin^8(x) = A^4, and  cos^8(x) = B^5

But as A is a fraction, A^4 < A.

Similarly A is a fraction, B^5 < B.

Therefore (A^4 + B^5) < (A + B)

But we know that A + B = sin^2(x) + cos^2(x) = 1

Therefore (A^4 + B^5) < 1

Therefore: sin^8(x) + cos^10(x) < 1