Prove that sin(180+x)=-sinx geometrically without using the expansion of sin(a+b). I have worked on it a lot but at the end every ends up up making a ray which makes an angle 'x' with the positive x-axis. Then this ray is revolved counterclockwise/anticlockwise so as to make the the angle made by the ray with the x-axis as 180+x. Then they say that as the first ray lies in the first quadrant there the sin of x is positive there because perpendicular and hypotenuse are both positive. For the second ray they say that it is symmetrically opposite to the first ray and so its base is negative, perpendicular is negative and hypotenuse is positive (this ray lies in the 3rd quadrant). So they say sin of this (180+x) is negative! But I fail to understand that how is this angle (180+x) contained in the triangle formed in the 3rd quadrant? The triangle that is made in the third quadrant by the revolved ray has an angle 'x' and not(180+x). Also as a matter of fact how can a triangle have an angle greater than 180 as its interior angle? It clearly violated the angle sum property of a triangle. That's why I need help in understanding this. Thanks
June 21st 2013, 09:59 PM
If you are using triangles, then you can't have angles greater than 180 degrees (in total and individually).
In order to have any individual angle greater than 180 you will need to have a quadrilateral or higher polygon if you decide to use this kind of method to prove your result.
June 21st 2013, 10:14 PM
Then what can be the possible proofs for this? Can you list them all down? I will look at them on my own.
June 21st 2013, 10:34 PM
The one off the top of my head that I can think of is to construct a geometric proof for sin(a+b) and then use that as a means to prove your result.
There are geometric ways to prove sin(a+b) and cos(a+b) but it's literally been more than a decade since I've seen them first hand.