[Math] If $\cos(-\theta) = \cos \theta$ then why is the value of $\cos(-\theta) $ negative when $\theta \lt -90^\circ$

trigonometry

I'm learning Trigonometry right now with myself and currently learning about trigonometric functions. I'm a little bit confused right now.

If $\cos(-\theta) = \cos \theta$, then why does $\cos \theta$ have negative values when $\theta$ is less than $-90^\circ$?

Best Answer

One of the ways of defining the trigonometric functions is geometrically.

See Wikipedia.

Unit Circle

Taking the Unit Circle (the circle of radius $1$ with center the origin, describable by the equation $x^2+y^2=1$), given a particular angle $\theta$, the value of cosine of $\theta$ is the $x$ coordinate for where the ray with angle $\theta$ above the positive $x$-axis intersects the circle.

Notice that for $90^\circ<\theta<180^\circ$ you will be in the top left quadrant, and in particular will have negative $x$ coordinates. Similarly for $-180^\circ<\theta<-90^\circ$ it will lie in the bottom left quadrant. In particular for something like $-100^\circ$.

(there are more complicated definitions which fix some inconsistencies with this definition as well as extend cosine to more abstract settings, but for now this definition is good enough)