What is the physical meaning of sine, cosine and tangent of an obtuse angle?
I have quite a few questions.
- First of all, for an angle $90^\circ\lt\theta\lt 180^\circ$, what would the sine/cosine/tangent of this angle be? What I'm saying is that a right angled triagle will always have all other angles acute. How is this possible?
- Why is only sine function positive in this quadrant while cosine and tangent are negative?
My teacher just told me to cram the values of trigonometric functions at different quadrants but I'm looking for a physical derivation for this. I'm also attaching a figure I drew to understand a right angled triangle with an obtuse angle.
2 Answers
$\begingroup$First of all, I upvoted your query re very positive approach, re
"but I'm looking for a physical derivation for this".
To understand this, in the realm of trigonometry,
where sine and cosine are functions of angles,
you need to consider the sine and cosine functions
against the backdrop of the unit circle.
Imagine a unit circle centered at the origin, that hits the $x$ and $y$
axes at points (1,0), (0,1), (-1,0), and (0, -1).
Consider any point in the unit circle that is in the first (upper right) quadrant. The point will have coordinates $(x,y).$
Let $\theta$ denote the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y).
Since the radius of the circle is 1, $\cos \theta = x$ and
$\sin \theta = y.$
Now imagine traveling around the arc of the unit circle until you reach the point (0,1).
This point may be construed to represent $90^{\circ}$, just as one complete revolution around the circle can be construed to represent $360^{\circ}.$
It is easy to see that $\cos(90^{\circ}) = 0$ and $\sin(90^{\circ}) = 1.$
Now imagine traveling around the arc to any point on the unit circle that is in the 2nd (upper left) quadrant.
Here, the point $(x,y)$ in the 2nd quadrant will have $x < 0$ and $y > 0.$
Again, just as before, consider $\theta$ to be the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y).
Here, by convention, $\cos \theta$ (again) $ = x$
and $\sin \theta$ (again) $ = y.$
Thus, it is easy to see that when $(x,y)$ is in the 2nd quadrant, and
$\theta$ is the angle formed by (0,0) -- (1,0) with (0,0) -- (x,y)
that $\cos \theta$ will by convention be $< 0$ and
$\sin \theta$ will by convention be $ > 0.$
A clear advantage of these conventions is that they facilitate the formulas shown at .
Another advantage of these conventions, which may not seem important in the realm of trigonometry, is that they facilitate the cosine and sine functions being continuous functions. This is a pandora's box that may not be worth exploring in the realm of trigonometry, but is still worth a very casual mention.
See .
$\endgroup$ $\begingroup$In obtuse angled triangle, you would draw a perpendicular. It may meet the third side externally.
Sine is perpendicular upon hypotenuse. In second quadrant, perpendicular (i.e. y-axis coordinate) is positive. So, sine is positive. Here, base (i.e x-axis coordinate) is negative. So, cosine and tangent are negative.
Is this clear now?
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