Given: 0 < a < pi/2
To prove: sin(a)^7 + cos(a)^7 < 1
Proof:
Since 0 < a < pi/2, sin(a) and cos(a) are both positive numbers less than 1.
For any positive number x less than 1, x^7 is also less than 1.
Therefore, sin(a)^7 and cos(a)^7 are both less than 1.
Adding two numbers less than 1 gives a number less than 2.
Therefore, sin(a)^7 + cos(a)^7 < 1.
Q.E.D.
Here is an explanation of the proof:
The first step is to show that sin(a) and cos(a) are both positive numbers less than 1. This is true because 0 < a < pi/2, which means that a is in the first quadrant. In the first quadrant, all trigonometric functions are positive.
The second step is to show that x^7 is less than 1 for any positive number x less than 1. This is true because as x gets closer to 0, x^7 gets closer to 0. Therefore, for any positive number x less than 1, x^7 is less than 1.
The third step is to add sin(a)^7 and cos(a)^7. Since both sin(a)^7 and cos(a)^7 are less than 1, their sum is also less than 1.
The fourth step is to conclude that sin(a)^7 + cos(a)^7 < 1. This is true because the sum of two numbers less than 1 is less than 2.
I hope this explanation is helpful. Please let me know if you have any other questions.
