For 0<x<pi/2, sin x and cos x are both less than 1 and greater than 0 (easy to see). We are also given that sin^2x+cos^2x=1. Use this to show that sin^7x+cos^7x<1 for 0<x<pi/2.

Unsure on how to proceed?

For the given domain,

since sinx < 1, sin^7x < sin^2x
same for cosx

so, sin^7 + cos^7 < sin^2 + cos^2 = 1

In general holds true for any power greater than 2.

Thanks man

To show that sin^7x + cos^7x < 1 for 0 < x < pi/2, we can use the fact that sin^2x + cos^2x = 1. We can work through the steps to derive the inequality.

1. Start with the expression sin^7x + cos^7x.

2. Rewrite sin^7x as (sin^2x)^3 * sinx, and rewrite cos^7x as (cos^2x)^3 * cosx. This is done to reuse the identity sin^2x + cos^2x = 1.

3. Substitute (sin^2x + cos^2x) for 1 in the previous step. Now we have:

(sin^2x)^3 * sinx + (cos^2x)^3 * cosx.

4. Notice that both sin^2x and cos^2x are between 0 and 1, according to the given information. Therefore, (sin^2x)^3 and (cos^2x)^3 are also between 0 and 1.

5. Since sinx and cosx are also between 0 and 1 (due to the given range 0 < x < pi/2), the entire expression (sin^2x)^3 * sinx + (cos^2x)^3 * cosx is the sum of positive values between 0 and 1.

6. Thus, the sum sin^7x + cos^7x is less than 1, as it is composed of values between 0 and 1 that sum to less than 1.

By following these steps, we have shown that sin^7x + cos^7x < 1 for 0 < x < pi/2.