Proove SinA + CosA > 1

Question

Steps to proove the above equation

Answer 1

Draw right triangle ABC with acute angle A and right angle C.  Let a = side opposite angle A, b = side opposite angle B and c = hypotenuse.

 

Then sinA = a/c and cosA = b/c

 

So, sinA + cosA = a/c + b/c = (a+b) / c.

 

In any triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

 

So, (a+b) / c > 1.

 

Therefore, if A is an acute angle, then sinA + cosA > 1.

 

The inequality is not valid for all angles, as Arturo illustrated in his answer. 

 

Answer 2

This cannot be proven because it is not generally true.  Here is a counterexample:

 

A = 90º

 

cos90º + sin90º = 0 + 1 = 1, which is not > 1

 

Here is another counterexample:

 

A = 0°

 

cos0º + sin0º = 1 + 0 = 1, which is not > 1

 

Perhaps you meant to ask a different question, such as SOLVE for A that makes the inequality true.