How do I prove that |sin z | >= |sinx|?

Question

I know that |sinz| ^2 = sin^2(x) + sinh^2(y)

But I have to prove that |sin z | >= |sinx|

I would normally make a good start on my quesitons before posting them but I don't even know the first step, so any initial help to get me on the right track would be nice... :-) My only thought is I must have to work with |sinz| ^2 and peform a sqaure root, but even then how do I know it's greater than? thanks guys

Laura

Answer 1

This actually in no way depends on any trigonometry.



For any c, c^2 >=0.

So for any b, b^2 + c^2 >= b^2 + 0.

So if a^2 = b^2 + c^2, a^2 >= b^2.

Take the square root of both sides and you're done!

Answer 2

try to start here:

if c² = a² + b², then | c | >= | a |

Answer 3

I HAVE NOT REACHED THAT LEVEL. THEREFORE I AM SORRY NOT TO GUIDE YOU.