# Prove that the inequality sin^8 (x)+cos^10 (x)<=1.

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### 3 Answers

Due to the relation 0=<sin ^2 (x)=<1, we'll have

0=<sin^8 (x)=<sin ^2(x) (1)

Also, due to the relation 0=<cos^(x)=<1, we'll have 0=<cos^10(x)=<cos^2(x) (2)

If we'll add the relations (1) and (2), we'll obtain

sin^8 (x) + cos^10(x)=< sin ^2(x) + cos^2(x)=1

Note that the symbol "=<", means "less or equal"!

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).

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

Toprove that sin^8x+cos^10X <= 1

We shall prove that an expression greater than or equal to the expressionon the left is less than or equal to1.

We know that: sin^2x+cos^2x=1 .

So, (sin^2x +cos^2x)^4=1^4=1 is an identity for all x.

Exapand the left side by using the binomial theorem and rearrange:

sin^8x+cos^8x+4((sin^2x)^3*cos^2x+sin^2x*(cos^2x)^3+6sin^4x*cos^4x = 1.

sin^8x+cos^8x+2sin^2xcos^2(2sin^4x+2cos^4x+3cos^2xcos^2x)

Therefore, sin^8x+cos^8x + (a positive quantity or 0) = 1.

So, sin^8x+cos^8x < =1 for all x .

But cos^8x > cos^10x , as 0 <=cos^2x<=1.

So, sin ^8x+cos^10x <= 1.

Hope this helps.