1. trig proof

Use what you learned in calculus to show that Sin(alpha)≤alpha and sin(alpha)/cos(alpha)≥alpha holds for all 0≤alpha≤pi/2

2. Hello,
Originally Posted by rmpatel5
Use what you learned in calculus to show that Sin(alpha)≤alpha and sin(alpha)/cos(alpha)≥alpha holds for all 0≤alpha≤pi/2
sin(alpha)≤alpha <--> sin(alpha)-alpha≤0
Take the derivative of sin(alpha)-alpha and see its variations between 0 and pi/2

Same goes for the other one.

3. Originally Posted by Moo
Hello,

sin(alpha)≤alpha <--> sin(alpha)-alpha≤0
Take the derivative of sin(alpha)-alpha and see its variations between 0 and pi/2

Same goes for the other one.
I took the der. of the first one and i get cos(alpha)≤1 and the second one sec^2(x)≥1

the first case:
alpha goes from 0 to pi/2

second case:
is just alpha is 0

i dont think my answer makes sense

4. You can use it using geometry.
Look at the picture.
The angle between the x-axis and the red line is $\displaystyle \alpha$.
While the circle is a unit circle.
Thus, the length of the arc made by $\displaystyle \alpha$ is $\displaystyle \alpha$.
While the length of the blue line is $\displaystyle \sin \alpha$.

Now reflect that image through the x-axis (second picture).
We see that the length of the blue line is $\displaystyle 2\sin \alpha$.
And the length of arc made by the blue line is $\displaystyle 2\alpha$.
Thus, $\displaystyle 2\sin \alpha \leq 2\alpha \implies \sin \alpha \leq \alpha$.
This is because the shortest distance between two points is a straight line.